Among the following the pair in which the two species are not isostructural is

Options

A. `IO_(3)^(-)` and `NH_(3)`
B. `BH_(4)^(-)` and `NH_(4)^(+)`
C. `PF_(6)^(-)` and `SF_(6)`
D. `SiF_(4)` and `SF_(4)`

Explanation

To determine which pair of species are not isostructural, we will analyze the molecular geometry and hybridization of each pair.

### Step 1: Analyze the first pair - IO3⁻ and NH3
- **IO3⁻ (Iodate ion)**:
- Iodine (I) has 7 valence electrons. In IO3⁻, it forms 3 sigma bonds with oxygen and has 1 lone pair.
- Total: 3 bond pairs + 1 lone pair = 4 electron pairs.
- Geometry: Tetrahedral (due to sp³ hybridization).

- **NH3 (Ammonia)**:
- Nitrogen (N) has 5 valence electrons. In NH3, it forms 3 sigma bonds with hydrogen and has 1 lone pair.
- Total: 3 bond pairs + 1 lone pair = 4 electron pairs.
- Geometry: Tetrahedral (due to sp³ hybridization).

**Conclusion**: IO3⁻ and NH3 are isostructural (both tetrahedral).

### Step 2: Analyze the second pair - BH4⁻ and NH4⁺
- **BH4⁻ (Boron tetrahydride)**:
- Boron (B) has 3 valence electrons. In BH4⁻, it forms 4 sigma bonds with hydrogen.
- Total: 4 bond pairs + 0 lone pairs = 4 electron pairs.
- Geometry: Tetrahedral (due to sp³ hybridization).

- **NH4⁺ (Ammonium ion)**:
- Nitrogen (N) has 5 valence electrons. In NH4⁺, it forms 4 sigma bonds with hydrogen.
- Total: 4 bond pairs + 0 lone pairs = 4 electron pairs.
- Geometry: Tetrahedral (due to sp³ hybridization).

**Conclusion**: BH4⁻ and NH4⁺ are isostructural (both tetrahedral).

### Step 3: Analyze the third pair - PF6⁻ and SF6
- **PF6⁻ (Phosphorus hexafluoride)**:
- Phosphorus (P) has 5 valence electrons. In PF6⁻, it forms 6 sigma bonds with fluorine.
- Total: 6 bond pairs + 0 lone pairs = 6 electron pairs.
- Geometry: Octahedral (due to sp³d² hybridization).

- **SF6 (Sulfur hexafluoride)**:
- Sulfur (S) has 6 valence electrons. In SF6, it forms 6 sigma bonds with fluorine.
- Total: 6 bond pairs + 0 lone pairs = 6 electron pairs.
- Geometry: Octahedral (due to sp³d² hybridization).

**Conclusion**: PF6⁻ and SF6 are isostructural (both octahedral).

### Step 4: Analyze the fourth pair - SiF4 and SF4
- **SiF4 (Silicon tetrafluoride)**:
- Silicon (Si) has 4 valence electrons. In SiF4, it forms 4 sigma bonds with fluorine.
- Total: 4 bond pairs + 0 lone pairs = 4 electron pairs.
- Geometry: Tetrahedral (due to sp³ hybridization).

- **SF4 (Sulfur tetrafluoride)**:
- Sulfur (S) has 6 valence electrons. In SF4, it forms 4 sigma bonds with fluorine and has 1 lone pair.
- Total: 4 bond pairs + 1 lone pair = 5 electron pairs.
- Geometry: Seesaw (due to sp³d hybridization).

**Conclusion**: SiF4 and SF4 are not isostructural (SiF4 is tetrahedral while SF4 is seesaw).

### Final Answer:
The pair in which the two species are not isostructural is **SiF4 and SF4**.