In solid `N_(2)O_(3)` and solid `N_(2)O_(5),N-O` hybridization in cationic part are respectively:

Options

A. `sp^(2)` and `sp`
B. `sp` and `sp^(2)`
C. `sp` and `sp`
D. none of these

Explanation

To determine the hybridization of the nitrogen-oxygen bonds in the cationic parts of solid \( N_2O_3 \) and solid \( N_2O_5 \), we will follow these steps:

### Step 1: Analyze the structure of \( N_2O_3 \)

1. **Identify the bonding in \( N_2O_3 \)**:
- In \( N_2O_3 \), one nitrogen atom is bonded to one oxygen atom via a double bond and to two other oxygen atoms via single bonds.
- The other nitrogen atom is bonded to one oxygen atom via a double bond and has two lone pairs of electrons.

2. **Count the charge centers**:
- Each nitrogen atom has three charge centers (one double bond and two single bonds for the first nitrogen, and one double bond and two lone pairs for the second nitrogen).

3. **Determine hybridization**:
- Since both nitrogen atoms have three charge centers, the hybridization for nitrogen in \( N_2O_3 \) is \( sp^2 \).

### Step 2: Analyze the structure of \( N_2O_5 \)

1. **Identify the cationic part**:
- In \( N_2O_5 \), the cationic part is \( NO_2^+ \).

2. **Count valence electrons**:
- The valence electrons of nitrogen (N) are 5.
- The number of surrounding atoms (excluding oxygen) is 2 (as there are two oxygen atoms bonded to nitrogen).
- The positive charge means we subtract 1 from the total.

3. **Apply the hybridization formula**:
- Hybridization = \(\frac{(Valence \, electrons + Number \, of \, surrounding \, atoms - Positive \, charge + Negative \, charge)}{2}\)
- Plugging in the values:
\[ \text{Hybridization} = \frac{(5 + 2 - 1 + 0)}{2} = \frac{6}{2} = 3 \]
- This indicates \( sp \) hybridization.

### Conclusion

- The hybridization of nitrogen in \( N_2O_3 \) is \( sp^2 \).
- The hybridization of nitrogen in \( N_2O_5 \) is \( sp \).

Thus, the final answer is:
- For \( N_2O_3 \): \( sp^2 \)
- For \( N_2O_5 \): \( sp \)