Match the following question
Column -I`" "` Column -II
(A) `XeO_(3), XeF_(2), ClF_(3), PCl_(5)` ( p ) All are `sp^(3) d` except one
( B `CH_(4) , XeO_(3) , ClO_(4)^(-) , ICl_(2)^(+)` (q) All are `sp^(3)` except one
( C ) ` BF_(4)^(-) , SO_(4)^(2-) , CO_(3)^(-) , POCl_(3)` ( r ) All are `sp^(3)`
( D ) `SO_(2) , SnCl_(2) ,SO_(3)^(-) , SO_(2) Cl_(2)`
`sp^(2)` and `sp^(3)` hybridisation
( t) All are `sp^(2)` except one

Answer

To solve the matching question, we need to determine the hybridization of each compound in Column I and then match them with the appropriate descriptions in Column II. Let's break it down step by step.

### Step 1: Determine the Hybridization of Each Compound

**(A) XeO₃, XeF₂, ClF₃, PCl₅**

1. **XeO₃**:
- Xenon has 8 valence electrons.
- In XeO₃, there are 3 double bonds with oxygen and 1 lone pair left.
- Hybridization: 3 (bond pairs) + 1 (lone pair) = 4 → **sp³**.

2. **XeF₂**:
- Xenon has 8 valence electrons.
- In XeF₂, there are 2 bonds with fluorine and 3 lone pairs left.
- Hybridization: 2 (bond pairs) + 3 (lone pairs) = 5 → **sp³d**.

3. **ClF₃**:
- Chlorine has 7 valence electrons.
- In ClF₃, there are 3 bonds with fluorine and 2 lone pairs left.
- Hybridization: 3 (bond pairs) + 2 (lone pairs) = 5 → **sp³d**.

4. **PCl₅**:
- Phosphorus has 5 valence electrons.
- In PCl₅, there are 5 bonds with chlorine and no lone pairs.
- Hybridization: 5 (bond pairs) = 5 → **sp³d**.

**Summary for (A)**:
- XeO₃: sp³
- XeF₂: sp³d
- ClF₃: sp³d
- PCl₅: sp³d

**Match**: All are sp³d except one (XeO₃). → **(A) → (p)**

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**(B) CH₄, XeO₃, ClO₄⁻, ICl₂⁺**

1. **CH₄**:
- Carbon has 4 valence electrons.
- In CH₄, there are 4 bonds with hydrogen and no lone pairs.
- Hybridization: 4 (bond pairs) = 4 → **sp³**.

2. **XeO₃**:
- As calculated above, XeO₃ is **sp³**.

3. **ClO₄⁻**:
- Chlorine has 7 valence electrons.
- In ClO₄⁻, there are 4 bonds with oxygen and no lone pairs.
- Hybridization: 4 (bond pairs) = 4 → **sp³**.

4. **ICl₂⁺**:
- Iodine has 7 valence electrons.
- In ICl₂⁺, there are 2 bonds with chlorine and 3 lone pairs.
- Hybridization: 2 (bond pairs) + 3 (lone pairs) = 5 → **sp³d**.

**Summary for (B)**:
- CH₄: sp³
- XeO₃: sp³
- ClO₄⁻: sp³
- ICl₂⁺: sp³d

**Match**: All are sp³ except one (ICl₂⁺). → **(B) → (q)**

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**(C) BF₄⁻, SO₄²⁻, CO₃²⁻, POCl₃**

1. **BF₄⁻**:
- Boron has 3 valence electrons.
- In BF₄⁻, there are 4 bonds with fluorine and no lone pairs.
- Hybridization: 4 (bond pairs) = 4 → **sp³**.

2. **SO₄²⁻**:
- Sulfur has 6 valence electrons.
- In SO₄²⁻, there are 4 bonds with oxygen and no lone pairs.
- Hybridization: 4 (bond pairs) = 4 → **sp³**.

3. **CO₃²⁻**:
- Carbon has 4 valence electrons.
- In CO₃²⁻, there are 3 bonds with oxygen and 1 lone pair.
- Hybridization: 3 (bond pairs) + 1 (lone pair) = 4 → **sp²**.

4. **POCl₃**:
- Phosphorus has 5 valence electrons.
- In POCl₃, there are 4 bonds with chlorine and no lone pairs.
- Hybridization: 4 (bond pairs) = 4 → **sp³**.

**Summary for (C)**:
- BF₄⁻: sp³
- SO₄²⁻: sp³
- CO₃²⁻: sp²
- POCl₃: sp³

**Match**: All are sp³ except one (CO₃²⁻). → **(C) → (q)**

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**(D) SO₂, SnCl₂, SO₃²⁻, SO₂Cl₂**

1. **SO₂**:
- Sulfur has 6 valence electrons.
- In SO₂, there are 2 bonds with oxygen and 1 lone pair.
- Hybridization: 2 (bond pairs) + 1 (lone pair) = 3 → **sp²**.

2. **SnCl₂**:
- Tin has 4 valence electrons.
- In SnCl₂, there are 2 bonds with chlorine and 2 lone pairs.
- Hybridization: 2 (bond pairs) + 2 (lone pairs) = 4 → **sp³**.

3. **SO₃²⁻**:
- Sulfur has 6 valence electrons.
- In SO₃²⁻, there are 3 bonds with oxygen and 1 lone pair.
- Hybridization: 3 (bond pairs) + 1 (lone pair) = 4 → **sp³**.

4. **SO₂Cl₂**:
- Similar to SO₂, with 2 bonds with oxygen and 2 bonds with chlorine, and 1 lone pair.
- Hybridization: 2 (bond pairs) + 1 (lone pair) = 3 → **sp²**.

**Summary for (D)**:
- SO₂: sp²
- SnCl₂: sp³
- SO₃²⁻: sp³
- SO₂Cl₂: sp²

**Match**: All are sp² except one (SnCl₂). → **(D) → (t)**

### Final Matches
- (A) → (p)
- (B) → (q)
- (C) → (r)
- (D) → (t)