Which of the following pair contains iso-structural species ?

Options

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

Explanation

To determine which pair contains iso-structural species, we need to analyze the hybridization and molecular geometry of each species provided in the options. Iso-structural species have the same type of bonding and geometry despite potentially differing in composition.

### Step-by-Step Solution:

1. **Identify the First Pair: CH3⁻ and C2H6**
- For CH3⁻:
- Valence electrons on carbon = 4
- Number of hydrogen atoms = 3
- Charge = -1
- Hybridization = (4 + 3 + 1) / 2 = 4 → SP³
- Geometry: 3 bond pairs + 1 lone pair = Pyramidal

- For C2H6:
- Valence electrons on carbon = 4 (for each carbon, total = 8)
- Number of hydrogen atoms = 6
- Charge = 0
- Hybridization = (8 + 6 + 0) / 2 = 7 → SP³
- Geometry: 6 bond pairs + 0 lone pairs = Tetrahedral

**Conclusion:** Not iso-structural.

2. **Identify the Second Pair: NH4⁺ and NH3**
- For NH4⁺:
- Valence electrons on nitrogen = 5
- Number of hydrogen atoms = 4
- Charge = +1
- Hybridization = (5 + 4 - 1) / 2 = 4 → SP³
- Geometry: 4 bond pairs + 0 lone pairs = Tetrahedral

- For NH3:
- Valence electrons on nitrogen = 5
- Number of hydrogen atoms = 3
- Charge = 0
- Hybridization = (5 + 3 + 0) / 2 = 4 → SP³
- Geometry: 3 bond pairs + 1 lone pair = Pyramidal

**Conclusion:** Not iso-structural.

3. **Identify the Third Pair: SF4 and BF4⁻**
- For SF4:
- Valence electrons on sulfur = 6
- Number of fluorine atoms = 4
- Charge = 0
- Hybridization = (6 + 4 + 0) / 2 = 5 → SP³D
- Geometry: 4 bond pairs + 1 lone pair = Seesaw

- For BF4⁻:
- Valence electrons on boron = 3
- Number of fluorine atoms = 4
- Charge = -1
- Hybridization = (3 + 4 + 1) / 2 = 4 → SP³
- Geometry: 4 bond pairs + 0 lone pairs = Tetrahedral

**Conclusion:** Not iso-structural.

4. **Identify the Fourth Pair: H2O and BeF2**
- For H2O:
- Valence electrons on oxygen = 6
- Number of hydrogen atoms = 2
- Charge = 0
- Hybridization = (6 + 2 + 0) / 2 = 4 → SP³
- Geometry: 2 bond pairs + 2 lone pairs = Bent

- For BeF2:
- Valence electrons on beryllium = 2
- Number of fluorine atoms = 2
- Charge = 0
- Hybridization = (2 + 2 + 0) / 2 = 2 → SP
- Geometry: 2 bond pairs + 0 lone pairs = Linear

**Conclusion:** Not iso-structural.

### Final Conclusion:
After analyzing all pairs, the only pair that contains iso-structural species is the third pair: **SF4 and BF4⁻**.