In which of the following pairs, both the species have the same hybridisation ?
(I) `SF_(4),XeF_(4) " " (II) I_(3)^(-),XeF_(2) " " (III) ICI_(4)^(-),SiCl_(4) " " (IV) ClO_(3)^(-),PO_(4)^(3-)`

Options

A. I,II
B. II, III
C. II, IV
D. I,II,III

Answer

To determine which pairs of species have the same hybridization, we will analyze each pair one by one, calculating the hybridization based on the number of bond pairs and lone pairs.

### Step-by-Step Solution:

**Pair I: SF4 and XeF4**
1. **SF4 (Sulfur Tetrafluoride)**:
- Sulfur (S) has 6 valence electrons.
- It forms 4 bonds with fluorine atoms.
- Total electrons = 6 (from S) + 4 (from 4 F) = 10.
- Number of lone pairs = 1 (since 4 bonds use 4 electrons).
- Steric number = Number of bond pairs + Number of lone pairs = 4 + 1 = 5.
- Hybridization = sp³d.

2. **XeF4 (Xenon Tetrafluoride)**:
- Xenon (Xe) has 8 valence electrons.
- It forms 4 bonds with fluorine atoms.
- Total electrons = 8 (from Xe) + 4 (from 4 F) = 12.
- Number of lone pairs = 2 (since 4 bonds use 4 electrons).
- Steric number = 4 + 2 = 6.
- Hybridization = sp³d².

**Conclusion for Pair I**: Different hybridizations (sp³d vs sp³d²).

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**Pair II: I3⁻ and XeF2**
1. **I3⁻ (Triiodide Ion)**:
- Iodine (I) has 7 valence electrons. With a -1 charge, it has 8 electrons.
- It forms 3 bonds with iodine atoms.
- Total electrons = 8 (from I) + 0 (from 3 I) = 8.
- Number of lone pairs = 2 (since 3 bonds use 3 electrons).
- Steric number = 3 + 2 = 5.
- Hybridization = sp³d.

2. **XeF2 (Xenon Difluoride)**:
- Xenon (Xe) has 8 valence electrons.
- It forms 2 bonds with fluorine atoms.
- Total electrons = 8 (from Xe) + 2 (from 2 F) = 10.
- Number of lone pairs = 3 (since 2 bonds use 2 electrons).
- Steric number = 2 + 3 = 5.
- Hybridization = sp³d.

**Conclusion for Pair II**: Same hybridization (sp³d).

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**Pair III: ICl4⁻ and SiCl4**
1. **ICl4⁻ (Iodine Tetrachloride Ion)**:
- Iodine (I) has 7 valence electrons. With a -1 charge, it has 8 electrons.
- It forms 4 bonds with chlorine atoms.
- Total electrons = 8 (from I) + 4 (from 4 Cl) = 12.
- Number of lone pairs = 2 (since 4 bonds use 4 electrons).
- Steric number = 4 + 2 = 6.
- Hybridization = sp³d².

2. **SiCl4 (Silicon Tetrachloride)**:
- Silicon (Si) has 4 valence electrons.
- It forms 4 bonds with chlorine atoms.
- Total electrons = 4 (from Si) + 4 (from 4 Cl) = 8.
- Number of lone pairs = 0.
- Steric number = 4 + 0 = 4.
- Hybridization = sp³.

**Conclusion for Pair III**: Different hybridizations (sp³d² vs sp³).

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**Pair IV: ClO3⁻ and PO4³⁻**
1. **ClO3⁻ (Chlorate Ion)**:
- Chlorine (Cl) has 7 valence electrons. With a -1 charge, it has 8 electrons.
- It forms 3 bonds with oxygen atoms.
- Total electrons = 8 (from Cl) + 3 (from 3 O) = 11.
- Number of lone pairs = 1 (since 3 bonds use 3 electrons).
- Steric number = 3 + 1 = 4.
- Hybridization = sp³.

2. **PO4³⁻ (Phosphate Ion)**:
- Phosphorus (P) has 5 valence electrons. With a -3 charge, it has 8 electrons.
- It forms 4 bonds with oxygen atoms.
- Total electrons = 5 (from P) + 4 (from 4 O) = 9.
- Number of lone pairs = 0.
- Steric number = 4 + 0 = 4.
- Hybridization = sp³.

**Conclusion for Pair IV**: Same hybridization (sp³).

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### Final Answer:
The pairs where both species have the same hybridization are:
- Pair II: I3⁻ and XeF2 (sp³d)
- Pair IV: ClO3⁻ and PO4³⁻ (sp³)