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

Options

A. `[NF_(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 molecule. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Isostructural Species
Isostructural species are those that have the same shape and hybridization. We will use the formula for hybridization to find out the hybridization of each species.

### Step 2: Use the Hybridization Formula
The formula for hybridization is:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of Valence Electrons of Central Atom} + M - C + A \right) \]
Where:
- \( M \) = number of monovalent atoms
- \( C \) = positive charge
- \( A \) = negative charge

### Step 3: Analyze Each Pair

#### Pair 1: NF3 and BF3
1. **NF3:**
- Valence electrons of N = 5
- M (F) = 3
- C = 0 (no positive charge)
- A = 0 (no negative charge)
- Hybridization = \( \frac{1}{2} (5 + 3 - 0 + 0) = \frac{1}{2} (8) = 4 \) → sp³
- Shape = Tetrahedral (or pyramidal due to lone pair)

2. **BF3:**
- Valence electrons of B = 3
- M (F) = 3
- C = 0
- A = 0
- Hybridization = \( \frac{1}{2} (3 + 3 - 0 + 0) = \frac{1}{2} (6) = 3 \) → sp²
- Shape = Trigonal planar

**Conclusion:** NF3 and BF3 are not isostructural.

#### Pair 2: BF4⁻ and NH4⁺
1. **BF4⁻:**
- Valence electrons of B = 3
- M (F) = 4
- C = 0
- A = 1 (negative charge)
- Hybridization = \( \frac{1}{2} (3 + 4 - 0 + 1) = \frac{1}{2} (8) = 4 \) → sp³
- Shape = Tetrahedral

2. **NH4⁺:**
- Valence electrons of N = 5
- M (H) = 4
- C = 1 (positive charge)
- A = 0
- Hybridization = \( \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} (8) = 4 \) → sp³
- Shape = Tetrahedral

**Conclusion:** BF4⁻ and NH4⁺ are isostructural.

#### Pair 3: BCl3 and BrCl3
1. **BCl3:**
- Valence electrons of B = 3
- M (Cl) = 3
- C = 0
- A = 0
- Hybridization = \( \frac{1}{2} (3 + 3 - 0 + 0) = \frac{1}{2} (6) = 3 \) → sp²
- Shape = Trigonal planar

2. **BrCl3:**
- Valence electrons of Br = 7
- M (Cl) = 3
- C = 0
- A = 0
- Hybridization = \( \frac{1}{2} (7 + 3 - 0 + 0) = \frac{1}{2} (10) = 5 \) → sp³d
- Shape = Trigonal bipyramidal

**Conclusion:** BCl3 and BrCl3 are not isostructural.

#### Pair 4: NH3 and NO3⁻
1. **NH3:**
- Valence electrons of N = 5
- M (H) = 3
- C = 0
- A = 0
- Hybridization = \( \frac{1}{2} (5 + 3 - 0 + 0) = \frac{1}{2} (8) = 4 \) → sp³
- Shape = Tetrahedral (or pyramidal due to lone pair)

2. **NO3⁻:**
- Valence electrons of N = 5
- M (O) = 3 (Oxygen is divalent)
- C = 0
- A = 1 (negative charge)
- Hybridization = \( \frac{1}{2} (5 + 3 - 0 + 1) = \frac{1}{2} (9) = 4.5 \) → sp² (resonance structure)
- Shape = Trigonal planar

**Conclusion:** NH3 and NO3⁻ are not isostructural.

### Final Conclusion
The only isostructural pair among the given options is **BF4⁻ and NH4⁺**.