In which pair both are not isostructural :-

Options

A. `BH_(4)^(-) & AlH_(4)^(-) `
B. `NH_(4)^(+) & PH_(4)^(+)`
C. `PCl_(6)^(-) & SiF_(6)^(-2)`
D. `BCl_(4)^(-) & Icl_(4)^(-)`

Answer

To determine which pair of compounds is not isostructural, we will analyze the hybridization of each compound in the given pairs. Isostructural compounds have the same hybridization and shape.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Charges:**
- Pair 1: BH₄⁻ and AlH₄⁻
- Pair 2: NH₄⁺ and PH₄⁺
- Pair 3: PCl₆⁻ and SiF₆²⁻
- Pair 4: BCl₄⁻ and ICl₄⁻

2. **Calculate Hybridization for Each Compound:**
- The formula for hybridization is:
\[ \text{Hybridization} = \frac{1}{2} (V + H - C + A) \]
where:
- \( V \) = number of valence electrons of the central atom
- \( H \) = number of monovalent atoms attached
- \( C \) = positive charge
- \( A \) = negative charge

3. **Calculate for BH₄⁻:**
- Valence electrons (B) = 3
- Monovalent atoms (H) = 4
- Charge = -1
\[ \text{Hybridization} = \frac{1}{2} (3 + 4 - 0 + 1) = \frac{1}{2} (8) = 4 \quad \text{(sp}^3\text{)} \]

4. **Calculate for AlH₄⁻:**
- Valence electrons (Al) = 3
- Monovalent atoms (H) = 4
- Charge = -1
\[ \text{Hybridization} = \frac{1}{2} (3 + 4 - 0 + 1) = \frac{1}{2} (8) = 4 \quad \text{(sp}^3\text{)} \]

5. **Conclusion for Pair 1:**
- Both BH₄⁻ and AlH₄⁻ have sp³ hybridization. They are isostructural.

6. **Calculate for NH₄⁺:**
- Valence electrons (N) = 5
- Monovalent atoms (H) = 4
- Charge = +1
\[ \text{Hybridization} = \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} (8) = 4 \quad \text{(sp}^3\text{)} \]

7. **Calculate for PH₄⁺:**
- Valence electrons (P) = 5
- Monovalent atoms (H) = 4
- Charge = +1
\[ \text{Hybridization} = \frac{1}{2} (5 + 4 - 1 + 0) = \frac{1}{2} (8) = 4 \quad \text{(sp}^3\text{)} \]

8. **Conclusion for Pair 2:**
- Both NH₄⁺ and PH₄⁺ have sp³ hybridization. They are isostructural.

9. **Calculate for PCl₆⁻:**
- Valence electrons (P) = 5
- Monovalent atoms (Cl) = 6
- Charge = -1
\[ \text{Hybridization} = \frac{1}{2} (5 + 6 - 0 + 1) = \frac{1}{2} (12) = 6 \quad \text{(sp}^3\text{d}^2\text{)} \]

10. **Calculate for SiF₆²⁻:**
- Valence electrons (Si) = 4
- Monovalent atoms (F) = 6
- Charge = -2
\[ \text{Hybridization} = \frac{1}{2} (4 + 6 - 0 + 2) = \frac{1}{2} (12) = 6 \quad \text{(sp}^3\text{d}^2\text{)} \]

11. **Conclusion for Pair 3:**
- Both PCl₆⁻ and SiF₆²⁻ have sp³d² hybridization. They are isostructural.

12. **Calculate for BCl₄⁻:**
- Valence electrons (B) = 3
- Monovalent atoms (Cl) = 4
- Charge = -1
\[ \text{Hybridization} = \frac{1}{2} (3 + 4 - 0 + 1) = \frac{1}{2} (8) = 4 \quad \text{(sp}^3\text{)} \]

13. **Calculate for ICl₄⁻:**
- Valence electrons (I) = 7
- Monovalent atoms (Cl) = 4
- Charge = -1
\[ \text{Hybridization} = \frac{1}{2} (7 + 4 - 0 + 1) = \frac{1}{2} (12) = 6 \quad \text{(sp}^3\text{d}^2\text{)} \]

14. **Conclusion for Pair 4:**
- BCl₄⁻ has sp³ hybridization, while ICl₄⁻ has sp³d² hybridization. They are **not isostructural**.

### Final Answer:
The pair that is not isostructural is **BCl₄⁻ and ICl₄⁻**.