In which of the following pairs, the two species are isostructural :

Options

A. `SO_(4)^(2-)and NO_(3)^(-)`
B. `BF_(3)and NF_(3)`
C. `BrO_(3)^(-)and XeO_(3)`
D. `SF_(4)and XeF_(4)`

Explanation

To determine which pair of species is isostructural, we need to analyze their hybridization and molecular geometry. Isostructural species have the same geometry, which can be deduced from their hybridization states.

### Step-by-Step Solution:

1. **Identify the Species**: We need to analyze the given pairs of species to find out their hybridization and molecular geometry.

2. **Calculate Hybridization for Each Species**:
- **Formula for Hybridization**:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons on central atom} + \text{Number of monovalent species} + \text{Charge} \right) \]

3. **Analyze the First Pair (SO4²⁻ and NO3⁻)**:
- **SO4²⁻**:
- Valence electrons on S = 6
- Charge = +2 (because of -2 charge)
- Hybridization:
\[ \frac{1}{2} (6 + 0 + 2) = \frac{8}{2} = 4 \quad \text{(SP³)} \]
- Geometry: Tetrahedral
- **NO3⁻**:
- Valence electrons on N = 5
- Charge = +1 (because of -1 charge)
- Hybridization:
\[ \frac{1}{2} (5 + 1 + 0) = \frac{6}{2} = 3 \quad \text{(SP²)} \]
- Geometry: Trigonal planar
- **Conclusion**: Not isostructural.

4. **Analyze the Second Pair (BF3 and NF3)**:
- **BF3**:
- Valence electrons on B = 3
- Charge = 0
- Hybridization:
\[ \frac{1}{2} (3 + 0 + 0) = \frac{3}{2} = 3 \quad \text{(SP²)} \]
- Geometry: Trigonal planar
- **NF3**:
- Valence electrons on N = 5
- Charge = 0
- Hybridization:
\[ \frac{1}{2} (5 + 3 + 0) = \frac{8}{2} = 4 \quad \text{(SP³)} \]
- Geometry: Tetrahedral
- **Conclusion**: Not isostructural.

5. **Analyze the Third Pair (PF3 and ClF3)**:
- **PF3**:
- Valence electrons on P = 5
- Charge = 0
- Hybridization:
\[ \frac{1}{2} (5 + 3 + 0) = \frac{8}{2} = 4 \quad \text{(SP³)} \]
- Geometry: Tetrahedral with 1 lone pair (Trigonal pyramidal)
- **ClF3**:
- Valence electrons on Cl = 7
- Charge = 0
- Hybridization:
\[ \frac{1}{2} (7 + 3 + 0) = \frac{10}{2} = 5 \quad \text{(SP³D)} \]
- Geometry: T-shaped
- **Conclusion**: Not isostructural.

6. **Analyze the Fourth Pair (XeF4 and XeF3)**:
- **XeF4**:
- Valence electrons on Xe = 8
- Charge = 0
- Hybridization:
\[ \frac{1}{2} (8 + 0 + 0) = \frac{8}{2} = 4 \quad \text{(SP³D²)} \]
- Geometry: Square planar
- **XeF3**:
- Valence electrons on Xe = 8
- Charge = 0
- Hybridization:
\[ \frac{1}{2} (8 + 3 + 0) = \frac{11}{2} = 5.5 \quad \text{(SP³D)} \]
- Geometry: T-shaped
- **Conclusion**: Not isostructural.

7. **Final Conclusion**: After analyzing all pairs, the only pair that exhibits the same hybridization and geometry is the third pair (PF3 and ClF3), which are both trigonal pyramidal.

### Final Answer:
The correct isostructural pair is **PF3 and ClF3**.