In which of the following pairs of molecules/ ions, the central atoms have `sp^2`-hybridization ?

Options

A. `BF_3 and NO_2^-`
B. `NH_2^(-) and H_2O`
C. `BF_3 and NH_2^-`
D. `NO_2^(-) and NH_3`

Explanation

To determine which pairs of molecules or ions have central atoms with `sp^2` hybridization, we will calculate the steric number for each central atom in the given pairs. The steric number can be calculated using the formula:

\[ \text{Steric Number} = \frac{\text{Number of valence electrons of central atom} + \text{Number of monovalent atoms attached} + \text{(if anionic charge, add 1; if cationic charge, subtract 1)}}{2} \]

### Step-by-Step Solution:

1. **Identify the pairs of molecules/ions**:
Let's assume the pairs are:
- Pair 1: BF3 and NO2^-
- Pair 2: NH2^- and H2O
- Pair 3: BF3 and NH3
- Pair 4: NO2^- and H2O

2. **Calculate the steric number for BF3**:
- Boron (B) has 3 valence electrons.
- It is attached to 3 fluorine atoms (monovalent).
- Steric number = (3 + 3 + 0) / 2 = 3.
- Hybridization = `sp^2` (since steric number 3 corresponds to `sp^2`).

3. **Calculate the steric number for NO2^-**:
- Nitrogen (N) has 5 valence electrons.
- It is attached to 2 oxygen atoms (monovalent).
- There is a negative charge (anion), so we add 1.
- Steric number = (5 + 2 + 1) / 2 = 4.
- Hybridization = `sp^3` (since steric number 4 corresponds to `sp^3`).

4. **Calculate the steric number for NH2^-**:
- Nitrogen (N) has 5 valence electrons.
- It is attached to 2 hydrogen atoms (monovalent).
- There is a negative charge (anion), so we add 1.
- Steric number = (5 + 2 + 1) / 2 = 4.
- Hybridization = `sp^3`.

5. **Calculate the steric number for H2O**:
- Oxygen (O) has 6 valence electrons.
- It is attached to 2 hydrogen atoms (monovalent).
- Steric number = (6 + 2 + 0) / 2 = 4.
- Hybridization = `sp^3`.

6. **Calculate the steric number for NH3**:
- Nitrogen (N) has 5 valence electrons.
- It is attached to 3 hydrogen atoms (monovalent).
- Steric number = (5 + 3 + 0) / 2 = 4.
- Hybridization = `sp^3`.

7. **Summarize the results**:
- BF3: `sp^2`
- NO2^-: `sp^3`
- NH2^-: `sp^3`
- H2O: `sp^3`
- NH3: `sp^3`

8. **Identify the pairs with `sp^2` hybridization**:
- The only pair with `sp^2` hybridization is **BF3**.

### Conclusion:
The correct answer is the pair **BF3 and NO2^-** has the central atom with `sp^2` hybridization.