The `ppi-ppi` bond is present in

Options

A. `XeO_(3)`
B. `SO_(4)^(2-)`
C. `SO_(2)`
D. All of these

Answer

To determine where the `p pi - p pi` bond is present among the given options (XeO3, SO4^2-, SO2), we will analyze each compound step by step.

### Step 1: Analyze XeO3
1. **Valence Electrons**: Xenon (Xe) has 8 valence electrons, and each oxygen (O) has 6. Therefore, for XeO3:
- Total valence electrons = 8 (from Xe) + 3 × 6 (from 3 O) = 26 valence electrons.

2. **Hybridization**:
- The structure of XeO3 involves 3 double bonds with oxygen and 1 lone pair on xenon.
- Number of sigma bonds = 3 (one for each Xe-O bond).
- Number of lone pairs = 1.
- Therefore, hybridization = sigma bonds + lone pairs = 3 + 1 = 4 (sp^3).

3. **Bonding**:
- The bonds formed are 3 sigma bonds and 3 pi bonds (from the double bonds).
- These pi bonds are formed from p orbitals of oxygen and d orbitals of xenon, resulting in p pi - d pi bonds.
- **Conclusion**: No p pi - p pi bonds are present in XeO3.

### Step 2: Analyze SO4^2-
1. **Valence Electrons**: Sulfur (S) has 6 valence electrons, and each oxygen has 6. For SO4^2-:
- Total valence electrons = 6 (from S) + 4 × 6 (from 4 O) + 2 (for 2 negative charges) = 32 valence electrons.

2. **Hybridization**:
- The structure of SO4^2- involves 4 sigma bonds (one for each S-O bond).
- There are no lone pairs on sulfur.
- Therefore, hybridization = 4 (sp^3).

3. **Bonding**:
- The bonds formed are 4 sigma bonds and 2 pi bonds (from the double bonds).
- These pi bonds are p pi - d pi bonds (sulfur's d orbitals and oxygen's p orbitals).
- **Conclusion**: No p pi - p pi bonds are present in SO4^2-.

### Step 3: Analyze SO2
1. **Valence Electrons**: Sulfur (S) has 6 valence electrons, and each oxygen has 6. For SO2:
- Total valence electrons = 6 (from S) + 2 × 6 (from 2 O) = 18 valence electrons.

2. **Hybridization**:
- The structure of SO2 involves 2 sigma bonds (one for each S-O bond) and 1 lone pair on sulfur.
- Therefore, hybridization = 2 (sigma bonds) + 1 (lone pair) = 3 (sp^2).

3. **Bonding**:
- The bonds formed are 2 sigma bonds and 2 pi bonds (one from each double bond).
- These pi bonds are formed from the p orbitals of sulfur and oxygen, resulting in p pi - p pi bonds.
- **Conclusion**: p pi - p pi bonds are present in SO2.

### Final Conclusion
- **XeO3**: No p pi - p pi bonds.
- **SO4^2-**: No p pi - p pi bonds.
- **SO2**: Has p pi - p pi bonds.

Thus, the `p pi - p pi` bond is present in **SO2**.