Which of the following molecule have both `Ppi-Ppi` and `Ppi-dpi` bonding ?

Options

A. `ClO_(2)^(+)`
B. `NO_(2)^(+)`
C. `SO_(3)^(2-)`
D. `ClO_(4)^(-)`

Answer

To determine which of the given molecules exhibit both Pπ-Pπ and Pπ-Dπ bonding, we can analyze each option step by step.

### Step 1: Analyze ClO2^+
1. **Structure**: ClO2^+ has a chlorine atom bonded to two oxygen atoms with a positive charge.
2. **Hybridization**: Chlorine undergoes sp² hybridization due to the presence of one lone pair and two bond pairs.
3. **Electron Configuration**: Chlorine has the configuration of Ne 3s² 3p⁵. To form two bonds, it requires an excited state configuration of 3s² 3p⁴ 3d¹.
4. **Bonding**: The unhybridized p orbitals (p_y) can overlap with the p orbitals of oxygen (Pπ-Pπ), and the d orbital (d_xy) can overlap with the p orbital of oxygen (Pπ-Dπ).
5. **Conclusion**: ClO2^+ exhibits both Pπ-Pπ and Pπ-Dπ bonding.

### Step 2: Analyze NO2^+
1. **Structure**: NO2^+ has a nitrogen atom bonded to two oxygen atoms with a positive charge.
2. **Hybridization**: Nitrogen undergoes sp hybridization.
3. **Electron Configuration**: Nitrogen has the configuration of He 2s² 2p³. It requires two bonds, leading to an excited state of 2s² 2p¹.
4. **Bonding**: The unhybridized p orbitals (p_y and p_z) overlap with the p orbitals of oxygen (Pπ-Pπ).
5. **Conclusion**: NO2^+ exhibits only Pπ-Pπ bonding.

### Step 3: Analyze SO3^2−
1. **Structure**: SO3^2− has a sulfur atom bonded to three oxygen atoms with two negative charges.
2. **Hybridization**: Sulfur undergoes sp³ hybridization.
3. **Electron Configuration**: Sulfur has the configuration of Ne 3s² 3p⁴. It requires three bond pairs, leading to an excited state of 3s² 3p³.
4. **Bonding**: The unhybridized d orbital overlaps with the p orbitals of oxygen (Dπ-Pπ).
5. **Conclusion**: SO3^2− exhibits only Dπ-Pπ bonding.

### Step 4: Analyze ClO4^−
1. **Structure**: ClO4^− has a chlorine atom bonded to four oxygen atoms with a negative charge.
2. **Hybridization**: Chlorine undergoes sp³ hybridization.
3. **Electron Configuration**: Chlorine has the configuration of Ne 3s² 3p⁵. It requires four bond pairs, leading to an excited state of 3s¹ 3p³ 3d³.
4. **Bonding**: The unhybridized d orbitals overlap with the p orbitals of oxygen (Dπ-Pπ).
5. **Conclusion**: ClO4^− exhibits only Dπ-Pπ bonding.

### Final Conclusion
Among the analyzed molecules, only **ClO2^+** exhibits both Pπ-Pπ and Pπ-Dπ bonding. Therefore, the correct answer is **ClO2^+**.