The hybridization of atomic orbitals of nitrogen is `NO_(2)^(+), NO_(3)^(-)`, and `NH_(4)^(+)` respectively 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)` rspectively
D. `sp^(2),sp^(3),and sp` respectively

Explanation

To determine the hybridization of the nitrogen atom in the given species \( NO_2^+ \), \( NO_3^- \), and \( NH_4^+ \), we will analyze each molecule step by step.

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

1. **Determine the Lewis structure**:
- Nitrogen (N) has 5 valence electrons, and each oxygen (O) has 6 valence electrons.
- In \( NO_2^+ \), we have a positive charge, which means we have one less electron to consider.
- Total valence electrons = \( 5 + 2 \times 6 - 1 = 16 \).
- The Lewis structure will have nitrogen in the center with two oxygen atoms bonded to it. One of the oxygen atoms will form a double bond with nitrogen, while the other will form a single bond.

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

3. **Determine hybridization**:
- With 2 bond pairs and no lone pairs, the hybridization is \( sp \) (linear geometry).

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

1. **Determine the Lewis structure**:
- For \( NO_3^- \), nitrogen has 5 valence electrons, and each oxygen has 6 valence electrons.
- The negative charge adds one more electron.
- Total valence electrons = \( 5 + 3 \times 6 + 1 = 24 \).
- The Lewis structure will have nitrogen in the center with three oxygen atoms bonded to it. One of the oxygen atoms will form a double bond with nitrogen, while the other two will form single bonds.

2. **Count the bond pairs and lone pairs**:
- There are 3 bond pairs and no lone pairs on nitrogen.

3. **Determine hybridization**:
- With 3 bond pairs and no lone pairs, the hybridization is \( sp^2 \) (trigonal planar geometry).

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

1. **Determine the Lewis structure**:
- Nitrogen has 5 valence electrons, and each hydrogen (H) has 1 valence electron.
- The positive charge means we have one less electron.
- Total valence electrons = \( 5 + 4 \times 1 - 1 = 8 \).
- The Lewis structure will have nitrogen in the center with four hydrogen atoms bonded to it.

2. **Count the bond pairs and lone pairs**:
- There are 4 bond pairs and no lone pairs on nitrogen.

3. **Determine hybridization**:
- With 4 bond pairs and no lone pairs, the hybridization is \( sp^3 \) (tetrahedral geometry).

### Summary of Hybridizations:
- \( NO_2^+ \): \( sp \)
- \( NO_3^- \): \( sp^2 \)
- \( NH_4^+ \): \( sp^3 \)