The hybridization of atomic orbitals of nitrogen in `NO_(2)^(+),NO_(3)^(-)"and" NH_(4)^(+)` respectively are

Options

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

Explanation

To determine the hybridization of atomic orbitals of nitrogen in the compounds \( NO_2^+, NO_3^-, \) and \( NH_4^+ \), we will analyze each compound step by step.

### Step 1: Hybridization in \( NO_2^+ \)

1. **Count Valence Electrons**:
- Nitrogen has 5 valence electrons.
- Each oxygen has 6 valence electrons, and there are 2 oxygen atoms: \( 6 \times 2 = 12 \).
- Since there is a positive charge, we subtract 1 electron: \( 5 + 12 - 1 = 16 \) total valence electrons.

2. **Draw the Lewis Structure**:
- Place nitrogen in the center and connect it to two oxygen atoms.
- Distribute the electrons to satisfy the octet rule.
- The structure will show nitrogen with 2 single bonds to oxygen and a positive charge.

3. **Determine the Geometry**:
- With 2 bond pairs and no lone pairs, the geometry is linear.

4. **Identify Hybridization**:
- For 2 bond pairs, the hybridization is \( sp \).

### Step 2: Hybridization in \( NO_3^- \)

1. **Count Valence Electrons**:
- Nitrogen has 5 valence electrons.
- Each oxygen has 6 valence electrons, and there are 3 oxygen atoms: \( 6 \times 3 = 18 \).
- Since there is a negative charge, we add 1 electron: \( 5 + 18 + 1 = 24 \) total valence electrons.

2. **Draw the Lewis Structure**:
- Place nitrogen in the center and connect it to three oxygen atoms.
- One of the bonds will be a double bond (to satisfy the octet rule), while the other two will be single bonds.
- Distribute the remaining electrons to the oxygen atoms.

3. **Determine the Geometry**:
- With 3 bond pairs and no lone pairs on nitrogen, the geometry is trigonal planar.

4. **Identify Hybridization**:
- For 3 bond pairs, the hybridization is \( sp^2 \).

### Step 3: Hybridization in \( NH_4^+ \)

1. **Count Valence Electrons**:
- Nitrogen has 5 valence electrons.
- Each hydrogen has 1 valence electron, and there are 4 hydrogen atoms: \( 1 \times 4 = 4 \).
- Since there is a positive charge, we subtract 1 electron: \( 5 + 4 - 1 = 8 \) total valence electrons.

2. **Draw the Lewis Structure**:
- Place nitrogen in the center and connect it to four hydrogen atoms.
- Each hydrogen atom forms a single bond with nitrogen.

3. **Determine the Geometry**:
- With 4 bond pairs and no lone pairs, the geometry is tetrahedral.

4. **Identify Hybridization**:
- For 4 bond pairs, the hybridization is \( sp^3 \).

### Final Answer:
- The hybridization of nitrogen in \( NO_2^+ \) is \( sp \).
- The hybridization of nitrogen in \( NO_3^- \) is \( sp^2 \).
- The hybridization of nitrogen in \( NH_4^+ \) is \( sp^3 \).