The hybridization of atomic orbitals of nitrogen in `N_(3)^(-),(H_(3)Si)_(3)N and (H_(3)C)_(3)N` are :

Options

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

Explanation

To determine the hybridization of nitrogen in the given compounds \( N_3^- \), \( (H_3Si)_3N \), and \( (H_3C)_3N \), we can follow these steps:

### Step 1: Analyze \( N_3^- \)
1. **Count Valence Electrons**: Each nitrogen has 5 valence electrons. For 3 nitrogen atoms, that gives us \( 3 \times 5 = 15 \) electrons. The negative charge adds 1 more electron, giving us a total of 16 electrons.
2. **Draw the Lewis Structure**:
- Arrange the nitrogen atoms in a linear configuration: \( N \equiv N - N \).
- Each nitrogen forms bonds: the terminal nitrogens form one bond with the central nitrogen, and the central nitrogen forms a triple bond with one of the terminal nitrogens.
3. **Determine Steric Number**:
- The central nitrogen has 2 sigma bonds (one with each terminal nitrogen) and no lone pairs. Thus, the steric number is \( 2 \).
4. **Identify Hybridization**:
- A steric number of 2 corresponds to \( sp \) hybridization.

### Step 2: Analyze \( (H_3Si)_3N \)
1. **Draw the Structure**: The nitrogen is bonded to three silicon atoms.
2. **Count Bonds and Lone Pairs**:
- Nitrogen forms 3 sigma bonds with silicon and has 1 lone pair.
3. **Determine Steric Number**:
- The steric number is \( 3 \) (from the 3 sigma bonds) + \( 1 \) (from the lone pair) = \( 4 \).
4. **Identify Hybridization**:
- A steric number of 4 corresponds to \( sp^3 \) hybridization. However, due to back-bonding with silicon's empty d-orbitals, the effective hybridization is \( sp^2 \).

### Step 3: Analyze \( (H_3C)_3N \)
1. **Draw the Structure**: The nitrogen is bonded to three methyl groups.
2. **Count Bonds and Lone Pairs**:
- Nitrogen forms 3 sigma bonds with the three carbon atoms and has 1 lone pair.
3. **Determine Steric Number**:
- The steric number is \( 3 \) (from the 3 sigma bonds) + \( 1 \) (from the lone pair) = \( 4 \).
4. **Identify Hybridization**:
- A steric number of 4 corresponds to \( sp^3 \) hybridization.

### Summary of Hybridizations
- For \( N_3^- \): Central nitrogen is \( sp \), terminal nitrogens are \( sp^2 \).
- For \( (H_3Si)_3N \): Nitrogen is \( sp^2 \) due to back-bonding.
- For \( (H_3C)_3N \): Nitrogen is \( sp^3 \).

### Final Answer
- The hybridizations of nitrogen in \( N_3^- \), \( (H_3Si)_3N \), and \( (H_3C)_3N \) are \( sp \) (central), \( sp^2 \), and \( sp^3 \) respectively.