The hybridisations of atomic orbitals of nitrogen in `NO_2, NO_3^(-) and NH_4` respectively are

Options

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

Explanation

To determine the hybridization of nitrogen in the compounds NO2, NO3^(-), and NH4, we can follow these steps:

### Step 1: Determine the Valency of Nitrogen and Other Atoms
- **For NO2**:
- Valency of Nitrogen (N) = 5
- Valency of Oxygen (O) = 6 (and there are 2 O atoms)
- Total = 5 + (2 × 6) = 5 + 12 = 17

- **For NO3^(-)**:
- Valency of Nitrogen (N) = 5
- Valency of Oxygen (O) = 6 (and there are 3 O atoms)
- There is a -1 charge, which means we add 1 electron.
- Total = 5 + (3 × 6) + 1 = 5 + 18 + 1 = 24

- **For NH4**:
- Valency of Nitrogen (N) = 5
- Valency of Hydrogen (H) = 1 (and there are 4 H atoms)
- Total = 5 + (4 × 1) = 5 + 4 = 9

### Step 2: Calculate the Effective Number of Electron Pairs
- For each total, we will divide by 8 to find the number of hybrid orbitals:
- **For NO2**: 17 / 8 = 2.125 (approximately 2 pairs)
- **For NO3^(-)**: 24 / 8 = 3 pairs
- **For NH4**: 9 / 8 = 1.125 (approximately 1 pair)

### Step 3: Determine the Hybridization
- The hybridization can be determined based on the number of effective pairs:
- **For NO2**: 2 pairs correspond to **sp² hybridization**.
- **For NO3^(-)**: 3 pairs correspond to **sp² hybridization**.
- **For NH4**: 1 pair corresponds to **sp³ hybridization**.

### Summary of Hybridization
- **NO2**: sp²
- **NO3^(-)**: sp²
- **NH4**: sp³