In which of the following molecules/ions in the central atom `sp^2`-hybridized?

Options

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

Explanation

To determine which of the given molecules or ions has a central atom that is `sp^2` hybridized, we will analyze each molecule step by step.

### Step-by-Step Solution:

1. **Identify the Central Atom**:
- For each molecule or ion, identify the central atom. The central atom is typically the least electronegative atom or the atom that can form the most bonds.

2. **Determine the Valence Electrons**:
- Count the number of valence electrons for the central atom. For example, nitrogen (N) has 5 valence electrons, oxygen (O) has 6, and boron (B) has 3.

3. **Account for Charge**:
- If the molecule or ion has a charge, adjust the number of valence electrons accordingly. For example, if there is a negative charge, add one electron to the total count.

4. **Count Bond Pairs and Lone Pairs**:
- Determine how many bonds the central atom forms with surrounding atoms (bond pairs) and how many lone pairs of electrons are left on the central atom.

5. **Determine Hybridization**:
- Use the formula for hybridization:
\[ \text{Hybridization} = \text{Number of Bond Pairs} + \text{Number of Lone Pairs} \]
- Based on the total, determine the hybridization type:
- 2 regions of electron density: `sp`
- 3 regions of electron density: `sp^2`
- 4 regions of electron density: `sp^3`

### Analysis of Each Pair:

1. **NO2⁻ (Nitrite ion)**:
- Central atom: Nitrogen (N)
- Valence electrons: 5 (N) + 1 (negative charge) = 6
- Bond pairs: 2 (with 2 O atoms)
- Lone pairs: 1
- Hybridization: \(2 \text{ (bond pairs)} + 1 \text{ (lone pair)} = 3 \rightarrow sp^2\)

2. **NH3⁻ (Ammonia ion)**:
- Central atom: Nitrogen (N)
- Valence electrons: 5 (N) + 1 (negative charge) = 6
- Bond pairs: 3 (with 3 H atoms)
- Lone pairs: 1
- Hybridization: \(3 \text{ (bond pairs)} + 1 \text{ (lone pair)} = 4 \rightarrow sp^3\)

3. **NH2⁻ (Amide ion)**:
- Central atom: Nitrogen (N)
- Valence electrons: 5 (N) + 1 (negative charge) = 6
- Bond pairs: 2 (with 2 H atoms)
- Lone pairs: 2
- Hybridization: \(2 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 4 \rightarrow sp^3\)

4. **H2O (Water)**:
- Central atom: Oxygen (O)
- Valence electrons: 6 (O)
- Bond pairs: 2 (with 2 H atoms)
- Lone pairs: 2
- Hybridization: \(2 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 4 \rightarrow sp^3\)

5. **BF3 (Boron trifluoride)**:
- Central atom: Boron (B)
- Valence electrons: 3 (B)
- Bond pairs: 3 (with 3 F atoms)
- Lone pairs: 0
- Hybridization: \(3 \text{ (bond pairs)} + 0 \text{ (lone pairs)} = 3 \rightarrow sp^2\)

6. **Conclusion**:
- The pairs that have `sp^2` hybridization are **NO2⁻** and **BF3**.

### Final Answer:
The molecules/ions with `sp^2` hybridization of the central atom are **NO2⁻ and BF3**.