In which of the following pairs, the two species are iso-structural ?
`{:((a)SO_(4)^(2-)and NO_(3)^(-),(b),BF_(3)andNF_(3)),((c )BrO_(3)^(-)and XeO_(3),(d),SF_(4)and XeF_(4)):}`

Options

A. `BrO_(3)^(-) and XeO_(3)`
B. `SF_(4) and XeF_(4)`
C. `SO_(3)^(2-) and NO_(3)^(-)`
D. `BF_(3) and NF_(3)`

Answer

To determine which pairs of species are iso-structural, we need to analyze their hybridization and molecular geometry. Let's go through each pair step by step.

### Step 1: Analyze SO₄²⁻ and NO₃⁻
1. **SO₄²⁻ (Sulfate Ion)**:
- Valence electrons: Sulfur (6) + 4 oxygen (4×6) - 2 (for the charge) = 6 + 24 - 2 = 28 electrons.
- Hybridization: Steric number = (number of atoms bonded + lone pairs) = 4 (4 oxygen) + 0 = 4.
- Hybridization type: sp³.
- Geometry: Tetrahedral.

2. **NO₃⁻ (Nitrate Ion)**:
- Valence electrons: Nitrogen (5) + 3 oxygen (3×6) - 1 (for the charge) = 5 + 18 - 1 = 22 electrons.
- Hybridization: Steric number = (number of atoms bonded + lone pairs) = 3 (3 oxygen) + 0 = 3.
- Hybridization type: sp².
- Geometry: Trigonal planar.

**Conclusion**: SO₄²⁻ and NO₃⁻ are not iso-structural.

### Step 2: Analyze BF₃ and NF₃
1. **BF₃ (Boron Trifluoride)**:
- Valence electrons: Boron (3) + 3 fluorine (3×7) = 3 + 21 = 24 electrons.
- Hybridization: Steric number = 3 (3 fluorine) + 0 = 3.
- Hybridization type: sp².
- Geometry: Trigonal planar.

2. **NF₃ (Nitrogen Trifluoride)**:
- Valence electrons: Nitrogen (5) + 3 fluorine (3×7) = 5 + 21 = 26 electrons.
- Hybridization: Steric number = 3 (3 fluorine) + 1 (lone pair) = 4.
- Hybridization type: sp³.
- Geometry: Pyramidal.

**Conclusion**: BF₃ and NF₃ are not iso-structural.

### Step 3: Analyze BrO₃⁻ and XeO₃
1. **BrO₃⁻ (Bromate Ion)**:
- Valence electrons: Bromine (7) + 3 oxygen (3×6) - 1 (for the charge) = 7 + 18 - 1 = 24 electrons.
- Hybridization: Steric number = 3 (3 oxygen) + 1 (lone pair) = 4.
- Hybridization type: sp³.
- Geometry: Pyramidal.

2. **XeO₃ (Xenon Trioxide)**:
- Valence electrons: Xenon (8) + 3 oxygen (3×6) = 8 + 18 = 26 electrons.
- Hybridization: Steric number = 3 (3 oxygen) + 1 (lone pair) = 4.
- Hybridization type: sp³.
- Geometry: Pyramidal.

**Conclusion**: BrO₃⁻ and XeO₃ are iso-structural.

### Step 4: Analyze SF₄ and XeF₄
1. **SF₄ (Sulfur Tetrafluoride)**:
- Valence electrons: Sulfur (6) + 4 fluorine (4×7) = 6 + 28 = 34 electrons.
- Hybridization: Steric number = 4 (4 fluorine) + 1 (lone pair) = 5.
- Hybridization type: sp³d.
- Geometry: Seesaw.

2. **XeF₄ (Xenon Tetrafluoride)**:
- Valence electrons: Xenon (8) + 4 fluorine (4×7) = 8 + 28 = 36 electrons.
- Hybridization: Steric number = 4 (4 fluorine) + 2 (lone pairs) = 6.
- Hybridization type: sp³d².
- Geometry: Square planar.

**Conclusion**: SF₄ and XeF₄ are not iso-structural.

### Final Conclusion
The only pair that is iso-structural is **(c) BrO₃⁻ and XeO₃**.