In which of the following molecular/ions `BF_(3),NO_(2)^(-),NH_(2)^(-)` and ` H_(2)O` the correct atom is `sp^(2)` hybridized ?

Options

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

Answer

To determine which of the given molecules or ions (BF₃, NO₂⁻, NH₂⁻, and H₂O) has an atom that is sp² hybridized, we can use the following systematic approach.

### Step-by-Step Solution:

1. **Identify the Central Atom**:
- For each molecule/ion, identify the central atom around which the hybridization will be determined.

2. **Count Valence Electrons**:
- Determine the number of valence electrons for the central atom.
- Use the periodic table to find the group number, which indicates the number of valence electrons.

3. **Count Monovalent Atoms**:
- Count the number of monovalent atoms (atoms that form one bond) attached to the central atom.

4. **Account for Charge**:
- If the species is an anion, add the charge to the total count. If it is a cation, subtract the charge.

5. **Apply the Hybridization Formula**:
- Use the formula:
\[ \text{Hybridization Index} = \frac{1}{2} \left( V + M + \text{(charge from anion)} - \text{(charge from cation)} \right) \]
- Where \( V \) is the number of valence electrons, \( M \) is the number of monovalent atoms, and the charge is adjusted based on whether it is an anion or cation.

6. **Determine Hybridization**:
- Based on the hybridization index calculated:
- If the index is 3, the hybridization is sp².
- If the index is 4, the hybridization is sp³.

### Detailed Analysis of Each Species:

1. **BF₃**:
- Central Atom: Boron (B)
- Valence Electrons (V): 3 (Group 13)
- Monovalent Atoms (M): 3 (3 Fluorine atoms)
- Charge: 0 (neutral)
- Calculation:
\[ \text{Hybridization Index} = \frac{1}{2} (3 + 3 + 0) = \frac{6}{2} = 3 \quad \text{(sp² hybridized)} \]

2. **NO₂⁻**:
- Central Atom: Nitrogen (N)
- Valence Electrons (V): 5 (Group 15)
- Monovalent Atoms (M): 2 (2 Oxygen atoms)
- Charge: -1 (anion)
- Calculation:
\[ \text{Hybridization Index} = \frac{1}{2} (5 + 2 + 1) = \frac{8}{2} = 4 \quad \text{(sp³ hybridized)} \]

3. **NH₂⁻**:
- Central Atom: Nitrogen (N)
- Valence Electrons (V): 5 (Group 15)
- Monovalent Atoms (M): 2 (2 Hydrogen atoms)
- Charge: -1 (anion)
- Calculation:
\[ \text{Hybridization Index} = \frac{1}{2} (5 + 2 + 1) = \frac{8}{2} = 4 \quad \text{(sp³ hybridized)} \]

4. **H₂O**:
- Central Atom: Oxygen (O)
- Valence Electrons (V): 6 (Group 16)
- Monovalent Atoms (M): 2 (2 Hydrogen atoms)
- Charge: 0 (neutral)
- Calculation:
\[ \text{Hybridization Index} = \frac{1}{2} (6 + 2 + 0) = \frac{8}{2} = 4 \quad \text{(sp³ hybridized)} \]

### Conclusion:
The species that has sp² hybridization is **BF₃**.