In the solid state, `N_(2)O_(5)` exists as `[NO_(2)]^(+)[NO_(3)]^(-)` . The hybridizations around the N atoms in `NO_(2)^(+)` and `NO_(3)^(-)` are, respectively :

Options

A. `sp and sp^(2)`
B. `sp and sp^(3)`
C. `sp^(2)` and `sp^(3)`
D. sp and sp

Explanation

To determine the hybridization around the nitrogen atoms in \( NO_2^+ \) and \( NO_3^- \), we will follow these steps:

### Step 1: Analyze the structure of \( NO_2^+ \)
1. **Count the valence electrons for nitrogen and oxygen**:
- Nitrogen (N) has 5 valence electrons.
- Each oxygen (O) has 6 valence electrons.
- For \( NO_2^+ \), we have one nitrogen and two oxygens, and we need to account for the positive charge, which means we subtract one electron from the total count.

Total valence electrons = \( 5 + 2 \times 6 - 1 = 16 \) electrons.

2. **Draw the Lewis structure**:
- Nitrogen forms a double bond with one oxygen and a single bond with the other oxygen.
- The structure can be represented as \( O=N^+=O \).

3. **Determine the number of bond pairs and lone pairs**:
- There are 2 bond pairs (one double bond and one single bond) and no lone pairs on nitrogen.

4. **Calculate the steric number**:
- Steric number = Number of bond pairs + Number of lone pairs = 2 + 0 = 2.

5. **Determine hybridization**:
- A steric number of 2 corresponds to \( sp \) hybridization.

### Step 2: Analyze the structure of \( NO_3^- \)
1. **Count the valence electrons for nitrogen and oxygen**:
- Nitrogen has 5 valence electrons.
- Each oxygen has 6 valence electrons.
- For \( NO_3^- \), we have one nitrogen and three oxygens, and we need to account for the negative charge, which means we add one electron to the total count.

Total valence electrons = \( 5 + 3 \times 6 + 1 = 24 \) electrons.

2. **Draw the Lewis structure**:
- Nitrogen forms one double bond with one oxygen and single bonds with the other two oxygens.
- The structure can be represented as \( O=N-O^- \) with one of the oxygens carrying a negative charge.

3. **Determine the number of bond pairs and lone pairs**:
- There are 3 bond pairs (one double bond and two single bonds) and no lone pairs on nitrogen.

4. **Calculate the steric number**:
- Steric number = Number of bond pairs + Number of lone pairs = 3 + 0 = 3.

5. **Determine hybridization**:
- A steric number of 3 corresponds to \( sp^2 \) hybridization.

### Final Answer
The hybridizations around the nitrogen atoms in \( NO_2^+ \) and \( NO_3^- \) are \( sp \) and \( sp^2 \), respectively.

---