In, `Cl_(2)O_(6)(l)+HF to P+Q`
If `H^(-)` of acid HF attraches with Q, then correct option of hybridization of Cl-atom and `angle OClO` in the P and Q ions:

Options

A. `P:sp^(2),gt120^(@)`
B. `Q:sp^(3):109^(@)28'`
C. `P:sp^(3), lt 109^(@)28'`
D. `Q:sp^(3),gt109^(@)28'`

Explanation

To solve the problem, we need to analyze the reaction between \( \text{Cl}_2\text{O}_6 \) and \( \text{HF} \) and determine the hybridization of the chlorine atom in the products \( P \) and \( Q \), as well as the angle \( \text{OClO} \).

### Step-by-Step Solution:

1. **Identify the Reaction Products:**
The reaction between \( \text{Cl}_2\text{O}_6 \) and \( \text{HF} \) yields two products:
- \( P = \text{ClO}_2 \)
- \( Q = \text{HClO}_4 \)

2. **Determine the Structure of \( P \) (ClO2):**
- In \( \text{ClO}_2 \), chlorine is the central atom bonded to two oxygen atoms.
- The structure can be represented as:
\[ \text{O} = \text{Cl} - \text{O} \]
- There is one lone pair on the chlorine atom.

3. **Calculate the Hybridization of \( P \) (ClO2):**
- The number of sigma bonds = 2 (one for each Cl-O bond).
- The number of lone pairs = 1.
- Hybridization can be calculated using the formula:
\[ \text{Hybridization} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 2 + 1 = 3 \]
- Therefore, the hybridization of chlorine in \( \text{ClO}_2 \) is \( \text{sp}^2 \).

4. **Determine the Angle \( \text{OClO} \) in \( P \):**
- The geometry of \( \text{ClO}_2 \) is bent due to the presence of the lone pair.
- The bond angle \( \text{OClO} \) is less than \( 120^\circ \) because of the lone pair repulsion.

5. **Determine the Structure of \( Q \) (HClO4):**
- In \( \text{HClO}_4 \), chlorine is bonded to four oxygen atoms.
- The structure can be represented as:
\[ \text{O} = \text{Cl} - \text{O} - \text{O} - \text{O} \]
- There are no lone pairs on the chlorine atom.

6. **Calculate the Hybridization of \( Q \) (HClO4):**
- The number of sigma bonds = 4 (one for each Cl-O bond).
- The number of lone pairs = 0.
- Hybridization can be calculated as:
\[ \text{Hybridization} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 4 + 0 = 4 \]
- Therefore, the hybridization of chlorine in \( \text{HClO}_4 \) is \( \text{sp}^3 \).

7. **Determine the Angle \( \text{OClO} \) in \( Q \):**
- The geometry of \( \text{HClO}_4 \) is tetrahedral.
- The bond angle \( \text{OClO} \) is \( 109.5^\circ \).

### Final Answers:
- Hybridization of Cl in \( P \) (ClO2) = \( \text{sp}^2 \)
- Angle \( \text{OClO} \) in \( P \) = less than \( 120^\circ \)
- Hybridization of Cl in \( Q \) (HClO4) = \( \text{sp}^3 \)
- Angle \( \text{OClO} \) in \( Q \) = \( 109.5^\circ \)