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 will analyze each compound step by step.

### Step 1: Determine the hybridization of nitrogen in \(N_3^-\)

1. **Draw the Lewis Structure**:
- The azide ion \(N_3^-\) consists of three nitrogen atoms. The structure can be represented as \(N \equiv N - N^-\), where the first nitrogen is triple bonded to the second nitrogen, and the second nitrogen is single bonded to the third nitrogen, which carries a negative charge.
- The central nitrogen atom has one lone pair.

2. **Count the Sigma Bonds**:
- There are 2 sigma bonds (one between the first and second nitrogen, and one between the second and third nitrogen).
- The lone pair on the central nitrogen does not count as a sigma bond.

3. **Determine Hybridization**:
- The total number of regions of electron density around the central nitrogen is 2 (from the sigma bonds).
- Therefore, the hybridization is \(sp\) (since 2 regions correspond to \(sp\) hybridization).

### Step 2: Determine the hybridization of nitrogen in \((H_3Si)_3N\)

1. **Draw the Lewis Structure**:
- The nitrogen is bonded to three \(SiH_3\) groups. Each \(SiH_3\) contributes one sigma bond to nitrogen.
- The nitrogen also has one lone pair.

2. **Count the Sigma Bonds**:
- There are 3 sigma bonds (one from each \(SiH_3\)).
- The lone pair on nitrogen counts as another region of electron density.

3. **Determine Hybridization**:
- The total number of regions of electron density around nitrogen is 4 (3 from sigma bonds and 1 from the lone pair).
- Therefore, the hybridization is \(sp^3\) (since 4 regions correspond to \(sp^3\) hybridization).

### Step 3: Determine the hybridization of nitrogen in \((H_3C)_3N\)

1. **Draw the Lewis Structure**:
- The nitrogen is bonded to three \(CH_3\) groups. Each \(CH_3\) contributes one sigma bond to nitrogen.
- The nitrogen also has one lone pair.

2. **Count the Sigma Bonds**:
- There are 3 sigma bonds (one from each \(CH_3\)).
- The lone pair on nitrogen counts as another region of electron density.

3. **Determine Hybridization**:
- The total number of regions of electron density around nitrogen is 4 (3 from sigma bonds and 1 from the lone pair).
- Therefore, the hybridization is \(sp^3\) (since 4 regions correspond to \(sp^3\) hybridization).

### Summary of Hybridizations
- For \(N_3^-\): Hybridization is \(sp\).
- For \((H_3Si)_3N\): Hybridization is \(sp^3\).
- For \((H_3C)_3N\): Hybridization is \(sp^3\).

### Final Answer
- The hybridizations of nitrogen in the compounds are:
- \(N_3^-\): \(sp\)
- \((H_3Si)_3N\): \(sp^3\)
- \((H_3C)_3N\): \(sp^3\)