Which of the following pairs are iso-structural ?

Options

A. `CH_(3)^(-)` and `CH_(3)^(+)`
B. `NH_(4)^(+)` and `BH_(4)^(-)`
C. `SO_(4)^(2-)` and `BF_(3)`
D. `NH_(2)^(-)` and `BeF_(2)`

Answer

To determine which of the given pairs are iso-structural, we need to analyze the hybridization and molecular geometry of each compound. Iso-structural compounds have the same shape or geometry.

### Step-by-Step Solution:

1. **Identify the Compounds:**
We have the following pairs to analyze:
- CH3⁻ and CH3⁺
- NH4⁺ and BH4⁻
- SO4²⁻ and BF3
- NH2⁻ and BeF2

2. **Calculate Hybridization and Geometry:**
We will use the formula:
\[ x = \frac{1}{2} \left( \text{Valence electrons of central atom} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]

- **For CH3⁻:**
- Valence electrons (C) = 4
- Monovalent atoms (H) = 3
- Charge = -1
\[ x = \frac{1}{2} (4 + 3 + 1) = \frac{8}{2} = 4 \quad \text{(sp³ hybridization, tetrahedral geometry)} \]

- **For CH3⁺:**
- Valence electrons (C) = 4
- Monovalent atoms (H) = 3
- Charge = +1
\[ x = \frac{1}{2} (4 + 3 - 1) = \frac{6}{2} = 3 \quad \text{(sp² hybridization, trigonal planar geometry)} \]

- **For NH4⁺:**
- Valence electrons (N) = 5
- Monovalent atoms (H) = 4
- Charge = +1
\[ x = \frac{1}{2} (5 + 4 - 1) = \frac{8}{2} = 4 \quad \text{(sp³ hybridization, tetrahedral geometry)} \]

- **For BH4⁻:**
- Valence electrons (B) = 3
- Monovalent atoms (H) = 4
- Charge = -1
\[ x = \frac{1}{2} (3 + 4 + 1) = \frac{8}{2} = 4 \quad \text{(sp³ hybridization, tetrahedral geometry)} \]

- **For SO4²⁻:**
- Valence electrons (S) = 6
- Monovalent atoms (O) = 0 (O is bivalent)
- Charge = -2
\[ x = \frac{1}{2} (6 + 0 + 2) = \frac{8}{2} = 4 \quad \text{(sp³ hybridization, tetrahedral geometry)} \]

- **For BF3:**
- Valence electrons (B) = 3
- Monovalent atoms (F) = 3
- Charge = 0
\[ x = \frac{1}{2} (3 + 3 + 0) = \frac{6}{2} = 3 \quad \text{(sp² hybridization, trigonal planar geometry)} \]

- **For NH2⁻:**
- Valence electrons (N) = 5
- Monovalent atoms (H) = 2
- Charge = +1
\[ x = \frac{1}{2} (5 + 2 - 1) = \frac{6}{2} = 3 \quad \text{(sp² hybridization, bent geometry)} \]

- **For BeF2:**
- Valence electrons (Be) = 2
- Monovalent atoms (F) = 2
- Charge = 0
\[ x = \frac{1}{2} (2 + 2 + 0) = \frac{4}{2} = 2 \quad \text{(sp hybridization, linear geometry)} \]

3. **Compare the Shapes:**
- **CH3⁻**: Tetrahedral
- **CH3⁺**: Trigonal planar
- **NH4⁺**: Tetrahedral
- **BH4⁻**: Tetrahedral
- **SO4²⁻**: Tetrahedral
- **BF3**: Trigonal planar
- **NH2⁻**: Bent
- **BeF2**: Linear

4. **Identify Iso-structural Pairs:**
The pairs that have the same geometry are:
- NH4⁺ and BH4⁻ (Both tetrahedral)

### Conclusion:
The correct answer is that the pair **NH4⁺ and BH4⁻** are iso-structural.