Some of the properties of the two species, `NO_(3)^(-)` and `H_(3)O^(+)` are described below.Which one of them is correct?

Options

A. Isostructural with same hybridization for the central atom
B. Isostructural with different hybridization for the central atom
C. Similar in hybridization for the central atom with different structures
D. Dissimilar in hybridization for the central atom with different structures.

Answer

To solve the question regarding the properties of the species \( \text{NO}_3^{-} \) and \( \text{H}_3\text{O}^{+} \), we need to analyze their hybridization and structures step by step.

### Step 1: Determine the Hybridization of \( \text{NO}_3^{-} \)

1. **Identify the central atom**: The central atom in \( \text{NO}_3^{-} \) is nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons.
3. **Count the attached atoms**: There are three oxygen atoms attached to nitrogen. Each oxygen is a divalent atom, but we only consider the number of monovalent atoms for hybridization calculations. Here, we treat oxygen as contributing to the overall structure.
4. **Add the ionic charge**: The negative charge on \( \text{NO}_3^{-} \) adds one more electron.
5. **Calculate the steric number**:
\[ \text{Steric number} = \frac{1}{2} \left( \text{(number of valence electrons on N)} + \text{(number of attached atoms)} + \text{(charge)} \right) \]
\[ = \frac{1}{2} (5 + 3 + 1) = \frac{9}{2} = 4.5 \text{ (which rounds to 3)} \]
Therefore, the steric number is 3, indicating \( \text{sp}^2 \) hybridization.

### Step 2: Determine the Structure of \( \text{NO}_3^{-} \)

- The structure of \( \text{NO}_3^{-} \) is trigonal planar due to \( \text{sp}^2 \) hybridization.

### Step 3: Determine the Hybridization of \( \text{H}_3\text{O}^{+} \)

1. **Identify the central atom**: The central atom in \( \text{H}_3\text{O}^{+} \) is oxygen (O).
2. **Count the valence electrons**: Oxygen has 6 valence electrons.
3. **Count the attached atoms**: There are three hydrogen atoms attached to oxygen.
4. **Subtract the ionic charge**: The positive charge on \( \text{H}_3\text{O}^{+} \) means we subtract one electron.
5. **Calculate the steric number**:
\[ \text{Steric number} = \frac{1}{2} \left( \text{(number of valence electrons on O)} + \text{(number of attached atoms)} - \text{(charge)} \right) \]
\[ = \frac{1}{2} (6 + 3 - 1) = \frac{8}{2} = 4 \]
Therefore, the steric number is 4, indicating \( \text{sp}^3 \) hybridization.

### Step 4: Determine the Structure of \( \text{H}_3\text{O}^{+} \)

- The structure of \( \text{H}_3\text{O}^{+} \) is pyramidal due to \( \text{sp}^3 \) hybridization.

### Step 5: Compare the Results

- **Hybridization**:
- \( \text{NO}_3^{-} \): \( \text{sp}^2 \)
- \( \text{H}_3\text{O}^{+} \): \( \text{sp}^3 \)
- **Structure**:
- \( \text{NO}_3^{-} \): Trigonal planar
- \( \text{H}_3\text{O}^{+} \): Pyramidal

### Conclusion

Based on the analysis:
- They have different hybridization and different structures.

Thus, the correct answer is: **Dissimilar hybridization for the central atom with different structures.**

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