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

Options

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

Explanation

To determine the hybridization of nitrogen in the molecules \( NO_2^+, NO_3^-, \) and \( NH_4^+ \), we can use the formula:

\[ \text{Number of hybrid orbitals} = \frac{1}{2} \left( \text{Number of valence electrons of central atom} + M - C + A \right) \]

Where:
- \( M \) = number of monovalent atoms attached to the central atom
- \( C \) = positive charge on the molecule
- \( A \) = negative charge on the molecule

Let's analyze each molecule step by step.

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

1. **Determine the number of valence electrons for nitrogen**: Nitrogen has 5 valence electrons.
2. **Identify the number of monovalent atoms (M)**: In \( NO_2^+ \), there are 0 monovalent atoms (oxygen is divalent).
3. **Identify the charge (C and A)**: The molecule has a positive charge (+1), so \( C = 1 \) and \( A = 0 \).

Using the formula:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} \left( 5 + 0 - 1 + 0 \right) = \frac{1}{2} \left( 4 \right) = 2 \]

Thus, nitrogen in \( NO_2^+ \) is **sp hybridized**.

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

1. **Determine the number of valence electrons for nitrogen**: Nitrogen has 5 valence electrons.
2. **Identify the number of monovalent atoms (M)**: In \( NO_3^- \), there are 0 monovalent atoms (oxygen is divalent).
3. **Identify the charge (C and A)**: The molecule has a negative charge (-1), so \( C = 0 \) and \( A = 1 \).

Using the formula:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} \left( 5 + 0 - 0 + 1 \right) = \frac{1}{2} \left( 6 \right) = 3 \]

Thus, nitrogen in \( NO_3^- \) is **sp² hybridized**.

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

1. **Determine the number of valence electrons for nitrogen**: Nitrogen has 5 valence electrons.
2. **Identify the number of monovalent atoms (M)**: In \( NH_4^+ \), there are 4 monovalent hydrogen atoms.
3. **Identify the charge (C and A)**: The molecule has a positive charge (+1), so \( C = 1 \) and \( A = 0 \).

Using the formula:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} \left( 5 + 4 - 1 + 0 \right) = \frac{1}{2} \left( 8 \right) = 4 \]

Thus, nitrogen in \( NH_4^+ \) is **sp³ hybridized**.

### Summary of Hybridizations:
- \( NO_2^+ \): sp hybridized
- \( NO_3^- \): sp² hybridized
- \( NH_4^+ \): sp³ hybridized

### Final Answer:
- \( NO_2^+ \) → sp
- \( NO_3^- \) → sp²
- \( NH_4^+ \) → sp³