In which of the following pairs, both the species have the same hybridisation ?
(I) `SF_(4),XeF_(4) " " (II) I_(3)^(-),XeF_(2) " " (III) ICI_(4)^(-),SiCl_(4) " " (IV) ClO_(3)^(-),PO_(4)^(3-)`

Options

A. I,II
B. II, III
C. II, IV
D. I,II,III

Answer

To determine which pairs of species have the same hybridization, we will analyze each pair one by one using the hybridization formula:

**Hybridization Formula:**
\[ \text{Hybridization number} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of single bonded atoms} + \text{Charge} \right) \]

### Step 1: Analyze Pair (I) SF₄ and XeF₄
1. **For SF₄:**
- Valence electrons of Sulfur (S) = 6
- Number of single bonded atoms (4 F) = 4
- Charge = 0 (neutral)
- Hybridization number = \( \frac{1}{2} (6 + 4 + 0) = \frac{10}{2} = 5 \)
- Hybridization = **sp³d**

2. **For XeF₄:**
- Valence electrons of Xenon (Xe) = 8
- Number of single bonded atoms (4 F) = 4
- Charge = 0 (neutral)
- Hybridization number = \( \frac{1}{2} (8 + 4 + 0) = \frac{12}{2} = 6 \)
- Hybridization = **sp³d²**

**Conclusion for Pair (I):** Different hybridizations.

### Step 2: Analyze Pair (II) I₃⁻ and XeF₂
1. **For I₃⁻:**
- Valence electrons of Iodine (I) = 7 (for the central I)
- Number of single bonded atoms = 2 (two I atoms)
- Charge = -1
- Hybridization number = \( \frac{1}{2} (7 + 2 - 1) = \frac{8}{2} = 4 \)
- Hybridization = **sp³**

2. **For XeF₂:**
- Valence electrons of Xenon (Xe) = 8
- Number of single bonded atoms (2 F) = 2
- Charge = 0 (neutral)
- Hybridization number = \( \frac{1}{2} (8 + 2 + 0) = \frac{10}{2} = 5 \)
- Hybridization = **sp³d**

**Conclusion for Pair (II):** Different hybridizations.

### Step 3: Analyze Pair (III) ICl₄⁻ and SiCl₄
1. **For ICl₄⁻:**
- Valence electrons of Iodine (I) = 7
- Number of single bonded atoms (4 Cl) = 4
- Charge = -1
- Hybridization number = \( \frac{1}{2} (7 + 4 - 1) = \frac{10}{2} = 5 \)
- Hybridization = **sp³d**

2. **For SiCl₄:**
- Valence electrons of Silicon (Si) = 4
- Number of single bonded atoms (4 Cl) = 4
- Charge = 0 (neutral)
- Hybridization number = \( \frac{1}{2} (4 + 4 + 0) = \frac{8}{2} = 4 \)
- Hybridization = **sp³**

**Conclusion for Pair (III):** Different hybridizations.

### Step 4: Analyze Pair (IV) ClO₃⁻ and PO₄³⁻
1. **For ClO₃⁻:**
- Valence electrons of Chlorine (Cl) = 7
- Number of single bonded atoms (3 O) = 3 (considering double bonds)
- Charge = -1
- Hybridization number = \( \frac{1}{2} (7 + 3 - 1) = \frac{9}{2} = 4 \)
- Hybridization = **sp³**

2. **For PO₄³⁻:**
- Valence electrons of Phosphorus (P) = 5
- Number of single bonded atoms (4 O) = 4 (considering double bonds)
- Charge = -3
- Hybridization number = \( \frac{1}{2} (5 + 4 - 3) = \frac{6}{2} = 3 \)
- Hybridization = **sp³**

**Conclusion for Pair (IV):** Same hybridizations.

### Final Conclusion:
The pairs with the same hybridization are:
- Pair (II): I₃⁻ and XeF₂ (sp³)
- Pair (IV): ClO₃⁻ and PO₄³⁻ (sp³)

### Answer:
The correct pairs are (II) and (IV).