Which of the following has different hybridization than other :-

Options

A. `I_(3)^(-)`
B. `XeO_(3)`
C. `SeCl_(4)`
D. `AsF_(4)^(-)`

Answer

To determine which of the given compounds has a different hybridization than the others, we will analyze the hybridization of each compound step by step.

### Step 1: Analyze I3⁻ (Triiodide Ion)
1. **Valence Electrons**: Iodine (I) has 7 valence electrons. For I3⁻, we have 3 iodine atoms and a -1 charge.
- Total valence electrons = (3 × 7) + 1 = 22 electrons.
2. **Hybridization Calculation**:
- Number of hybrid orbitals = (22 electrons) / 2 = 11.
- Hybridization = sp³d² (since 11 corresponds to 5 hybrid orbitals).

### Step 2: Analyze XeO3 (Xenon Trioxide)
1. **Valence Electrons**: Xenon (Xe) has 8 valence electrons. There are no monovalent atoms.
- Total valence electrons = 8.
2. **Hybridization Calculation**:
- Number of hybrid orbitals = (8 electrons) / 2 = 4.
- Hybridization = sp³ (since 4 corresponds to 4 hybrid orbitals).

### Step 3: Analyze SeCl4 (Selenium Tetrachloride)
1. **Valence Electrons**: Selenium (Se) has 6 valence electrons, and there are 4 chlorine (Cl) atoms.
- Total valence electrons = 6 + (4 × 7) = 34 electrons.
2. **Hybridization Calculation**:
- Number of hybrid orbitals = (34 electrons) / 2 = 17.
- Hybridization = sp³d (since 5 corresponds to 5 hybrid orbitals).

### Step 4: Analyze AsF4⁻ (Arsenic Tetrafluoride Ion)
1. **Valence Electrons**: Arsenic (As) has 5 valence electrons, and there are 4 fluorine (F) atoms, plus a -1 charge.
- Total valence electrons = 5 + (4 × 7) + 1 = 34 electrons.
2. **Hybridization Calculation**:
- Number of hybrid orbitals = (34 electrons) / 2 = 17.
- Hybridization = sp³d (since 5 corresponds to 5 hybrid orbitals).

### Conclusion
After analyzing all compounds:
- I3⁻: sp³d²
- XeO3: sp³
- SeCl4: sp³d
- AsF4⁻: sp³d

**The compound with different hybridization is XeO3 (option B).**