Which of the following species shows `ppi - dpi` bonding ?

Options

A. `BO_3^(3-)`
B. `NO_3^-`
C. `CO_3^(-2)`
D. `SO_3`

Answer

To determine which of the following species shows Pπ-Dπ bonding, we need to analyze the bonding characteristics of each option provided.

### Step-by-Step Solution:

1. **Understanding Pπ-Dπ Bonding**:
Pπ-Dπ bonding occurs when a p-orbital from one atom overlaps with a d-orbital from another atom. This type of bonding is typically seen in compounds where one of the atoms is from the third period or below in the periodic table, as these elements have accessible d-orbitals.

2. **Analyzing the Options**:
- **Option A: Boron Compounds**: Boron is in the second period and does not have d-orbitals. Therefore, it cannot participate in Pπ-Dπ bonding.
- **Option B: Nitrogen Compounds**: Nitrogen is also in the second period and lacks d-orbitals, ruling out Pπ-Dπ bonding.
- **Option C: Carbon Compounds**: Carbon, like boron and nitrogen, is in the second period and does not have d-orbitals, so it cannot form Pπ-Dπ bonds.
- **Option D: SO3 (Sulfur Trioxide)**: Sulfur is in the third period and has accessible d-orbitals. In SO3, sulfur can participate in Pπ-Dπ bonding with the oxygen atoms.

3. **Conclusion**:
The only species that shows Pπ-Dπ bonding is **SO3**. The sulfur atom can use its d-orbitals to form bonds with the oxygen atoms, resulting in Pπ-Dπ bonding.

### Final Answer:
The correct answer is **D: SO3**.