In which of the following pairs of molecules central atom is `sp^(2)` hybridized ?

Options

A. `NO_(2)^(-)and NH_(3)`
B. `BF_(3)and NH_(2)^(-)`
C. `NH_(2)^(-)and H_(2)O`
D. `BF_(3)and NH_(2)^(-)`

Explanation

To determine which pairs of molecules have a central atom that is `sp²` hybridized, we can follow these steps:

### Step 1: Understand the Hybridization Concept
Hybridization is a concept used to explain the bonding in molecules. The type of hybridization depends on the steric number, which is calculated using the formula:
\[ \text{Steric Number} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} + \text{Charge} \right) \]
Where:
- Valence electrons are the electrons in the outermost shell of the central atom.
- Monovalent atoms are atoms that form one bond (like H, F, Cl).
- Charge is added if the molecule has a negative charge and subtracted if it has a positive charge.

### Step 2: Analyze Each Molecule
Let's analyze the given pairs of molecules one by one.

#### Pair 1: NO₂⁻
1. **Valence Electrons**: Nitrogen (N) has 5 valence electrons.
2. **Monovalent Atoms**: There are 2 oxygen atoms (O), which are divalent, but we only consider the central atom's bonds, so we count 0 for monovalent.
3. **Charge**: The negative charge contributes +1.
4. **Calculation**:
\[ \text{Steric Number} = \frac{1}{2} (5 + 0 + 1) = \frac{1}{2} (6) = 3 \]
Since the steric number is 3, the hybridization is `sp²`.

#### Pair 2: NH₃
1. **Valence Electrons**: Nitrogen has 5 valence electrons.
2. **Monovalent Atoms**: There are 3 hydrogen atoms (H).
3. **Charge**: There is no charge.
4. **Calculation**:
\[ \text{Steric Number} = \frac{1}{2} (5 + 3 + 0) = \frac{1}{2} (8) = 4 \]
Since the steric number is 4, the hybridization is `sp³`.

#### Pair 3: BF₃
1. **Valence Electrons**: Boron (B) has 3 valence electrons.
2. **Monovalent Atoms**: There are 3 fluorine atoms (F).
3. **Charge**: There is no charge.
4. **Calculation**:
\[ \text{Steric Number} = \frac{1}{2} (3 + 3 + 0) = \frac{1}{2} (6) = 3 \]
Since the steric number is 3, the hybridization is `sp²`.

#### Pair 4: NH₂⁻
1. **Valence Electrons**: Nitrogen has 5 valence electrons.
2. **Monovalent Atoms**: There are 2 hydrogen atoms (H).
3. **Charge**: The negative charge contributes +1.
4. **Calculation**:
\[ \text{Steric Number} = \frac{1}{2} (5 + 2 + 1) = \frac{1}{2} (8) = 4 \]
Since the steric number is 4, the hybridization is `sp³`.

### Step 3: Conclusion
From the analysis:
- NO₂⁻ has `sp²` hybridization.
- BF₃ has `sp²` hybridization.
- NH₃ has `sp³` hybridization.
- NH₂⁻ has `sp³` hybridization.

Thus, the pairs of molecules where the central atom is `sp²` hybridized are **NO₂⁻ and BF₃**.