Isostructrual species are those which have the same shape and hybridisation. Among the given identify the isostructural pairs.

Options

A. `NH_(3)` and `BF_(3)`
B. `BF_(4)^(-)` and `NH_(4)^(+)`
C. `BCl_(3)` and `BrCl_(3)`
D. `NH_(3)` and `NO_(3)^(-)`

Explanation

To identify the isostructural pairs among the given species, we need to determine the hybridization and shape of each species. Isostructural species have the same hybridization and geometric shape.

### Step-by-Step Solution:

1. **Identify the Central Atom and Count Valence Electrons**:
- For each species, identify the central atom and count its valence electrons.

2. **Calculate the Steric Number**:
- Use the formula:
\[ \text{Steric Number} = \frac{1}{2} \left( V + M + C - A \right) \]
Where:
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms attached
- \( C \) = charge of the anion (if applicable)
- \( A \) = charge of the cation (if applicable)

3. **Determine Hybridization**:
- Based on the steric number, determine the hybridization:
- Steric number 2: sp
- Steric number 3: sp²
- Steric number 4: sp³
- Steric number 5: sp³d
- Steric number 6: sp³d²

4. **Determine Molecular Shape**:
- Use VSEPR theory to determine the shape based on the number of bond pairs and lone pairs.

5. **Compare Hybridization and Shape**:
- Identify pairs that have the same hybridization and shape.

### Application to Given Species:

1. **Ammonia (NH₃)**:
- Central atom: Nitrogen (5 valence electrons)
- Monovalent atoms: 3 (H)
- Charge: 0
- Steric Number: \( \frac{1}{2}(5 + 3 + 0 - 0) = 4 \) → Hybridization: sp³
- Shape: Pyramidal (1 lone pair, 3 bond pairs)

2. **Boron Trifluoride (BF₃)**:
- Central atom: Boron (3 valence electrons)
- Monovalent atoms: 3 (F)
- Charge: 0
- Steric Number: \( \frac{1}{2}(3 + 3 + 0 - 0) = 3 \) → Hybridization: sp²
- Shape: Trigonal planar

3. **Tetrafluoroborate Ion (BF₄⁻)**:
- Central atom: Boron (3 valence electrons)
- Monovalent atoms: 4 (F)
- Charge: -1
- Steric Number: \( \frac{1}{2}(3 + 4 + 1 - 0) = 4 \) → Hybridization: sp³
- Shape: Tetrahedral

4. **Ammonium Ion (NH₄⁺)**:
- Central atom: Nitrogen (5 valence electrons)
- Monovalent atoms: 4 (H)
- Charge: +1
- Steric Number: \( \frac{1}{2}(5 + 4 + 0 - 1) = 4 \) → Hybridization: sp³
- Shape: Tetrahedral

5. **Boron Trichloride (BCl₃)**:
- Central atom: Boron (3 valence electrons)
- Monovalent atoms: 3 (Cl)
- Charge: 0
- Steric Number: \( \frac{1}{2}(3 + 3 + 0 - 0) = 3 \) → Hybridization: sp²
- Shape: Trigonal planar

6. **Bromine Trichloride (BrCl₃)**:
- Central atom: Bromine (7 valence electrons)
- Monovalent atoms: 3 (Cl)
- Charge: 0
- Steric Number: \( \frac{1}{2}(7 + 3 + 0 - 0) = 5 \) → Hybridization: sp³d
- Shape: T-shaped (3 bond pairs, 2 lone pairs)

7. **Nitrate Ion (NO₃⁻)**:
- Central atom: Nitrogen (5 valence electrons)
- Monovalent atoms: 3 (O)
- Charge: -1
- Steric Number: \( \frac{1}{2}(5 + 3 + 1 - 0) = 4 \) → Hybridization: sp²
- Shape: Trigonal planar

### Conclusion:
- The only isostructural pair identified is **Tetrafluoroborate Ion (BF₄⁻)** and **Ammonium Ion (NH₄⁺)**, both having sp³ hybridization and tetrahedral shape.

### Final Answer:
The isostructural pairs are **BF₄⁻ and NH₄⁺**.