Which of the following molecule exhibit `P_(pi)-P_(pi)` bonding?

Options

A. `CO_(3)^(-2)`
B. `NO_(3)^(-1)`
C. `SiO_(2)`
D. (1) & (2) both

Explanation

To determine which of the given molecules exhibit \( P_{\pi}-P_{\pi} \) bonding, we will analyze the hybridization and bonding characteristics of each molecule mentioned in the question.

### Step-by-Step Solution:

1. **Identify the Molecules**: The question mentions three molecules: Carbonate (\( CO_3^{2-} \)), Nitrate (\( NO_3^{-} \)), and Silicon Dioxide (\( SiO_2 \)). We need to analyze each one for \( P_{\pi}-P_{\pi} \) bonding.

2. **Analyze Carbonate (\( CO_3^{2-} \))**:
- **Valence Electrons**: Carbon has 4 valence electrons, and each oxygen has 6. The total for \( CO_3^{2-} \) is:
\[ 4 + (3 \times 6) + 2 = 4 + 18 + 2 = 24 \text{ valence electrons} \]
- **Hybridization**: The central carbon atom has 3 bonding regions (3 oxygen atoms) and no lone pairs, leading to \( sp^2 \) hybridization.
- **Bonding**: The structure has one double bond (which involves \( P_{\pi}-P_{\sigma} \) bonding) and two single bonds (which are \( P_{\sigma}-P_{\sigma} \)). The remaining \( P \) orbital can participate in \( P_{\pi}-P_{\pi} \) bonding.
- **Conclusion**: Carbonate exhibits \( P_{\pi}-P_{\pi} \) bonding.

3. **Analyze Nitrate (\( NO_3^{-} \))**:
- **Valence Electrons**: Nitrogen has 5 valence electrons, and each oxygen has 6. The total for \( NO_3^{-} \) is:
\[ 5 + (3 \times 6) + 1 = 5 + 18 + 1 = 24 \text{ valence electrons} \]
- **Hybridization**: The central nitrogen atom has 3 bonding regions (3 oxygen atoms) and no lone pairs, leading to \( sp^2 \) hybridization.
- **Bonding**: The structure has one double bond and two single bonds. The remaining \( P \) orbital can also participate in \( P_{\pi}-P_{\pi} \) bonding.
- **Conclusion**: Nitrate exhibits \( P_{\pi}-P_{\pi} \) bonding.

4. **Analyze Silicon Dioxide (\( SiO_2 \))**:
- **Valence Electrons**: Silicon has 4 valence electrons, and each oxygen has 6. The total for \( SiO_2 \) is:
\[ 4 + (2 \times 6) = 4 + 12 = 16 \text{ valence electrons} \]
- **Hybridization**: Silicon in \( SiO_2 \) has \( sp \) hybridization due to the formation of two sigma bonds with oxygen.
- **Bonding**: Silicon does not form stable double bonds with oxygen due to poor overlap between the large silicon orbitals and the smaller oxygen orbitals.
- **Conclusion**: Silicon Dioxide does not exhibit \( P_{\pi}-P_{\pi} \) bonding.

5. **Final Answer**: The molecules that exhibit \( P_{\pi}-P_{\pi} \) bonding are Carbonate (\( CO_3^{2-} \)) and Nitrate (\( NO_3^{-} \)). Therefore, the answer is:
\[ \text{Option D: Carbonate and Nitrate} \]