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

Options

A. `NO_(2)^(-) and NH_3`
B. `BF_3 and NO_(2)^(-)`
C. `NH_(2)^(-) and H_2 O`
D. `BF_3 and NH_(2)^(-)`

Explanation

To determine which pairs of molecules/ions have a central atom with `sp^2` hybridization, we can use the hybridization formula:

\[ \text{Number of hybrid orbitals} = \frac{V + M - C + A}{2} \]

Where:
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms (like H, F, Cl, etc.)
- \( C \) = positive charge (if cationic)
- \( A \) = negative charge (if anionic)

Let’s analyze each molecule/ion step by step:

### Step 1: Analyze NO2⁻
- **Valence electrons of Nitrogen (N)**: 5
- **Monovalent atoms (Oxygen)**: 2 (since there are 2 O atoms)
- **Anionic charge**: 1 (since it’s NO2⁻)

Using the formula:
\[ \text{Hybrid orbitals} = \frac{5 + 2 + 1}{2} = \frac{8}{2} = 4 \]
This indicates that NO2⁻ has `sp^2` hybridization.

### Step 2: Analyze NH3
- **Valence electrons of Nitrogen (N)**: 5
- **Monovalent atoms (Hydrogen)**: 3 (since there are 3 H atoms)
- **Charge**: 0

Using the formula:
\[ \text{Hybrid orbitals} = \frac{5 + 3 + 0}{2} = \frac{8}{2} = 4 \]
This indicates that NH3 has `sp^3` hybridization.

### Step 3: Analyze BF3
- **Valence electrons of Boron (B)**: 3
- **Monovalent atoms (Fluorine)**: 3 (since there are 3 F atoms)
- **Charge**: 0

Using the formula:
\[ \text{Hybrid orbitals} = \frac{3 + 3 + 0}{2} = \frac{6}{2} = 3 \]
This indicates that BF3 has `sp^2` hybridization.

### Step 4: Analyze NH2⁻
- **Valence electrons of Nitrogen (N)**: 5
- **Monovalent atoms (Hydrogen)**: 2 (since there are 2 H atoms)
- **Anionic charge**: 1 (since it’s NH2⁻)

Using the formula:
\[ \text{Hybrid orbitals} = \frac{5 + 2 + 1}{2} = \frac{8}{2} = 4 \]
This indicates that NH2⁻ has `sp^3` hybridization.

### Step 5: Analyze H2O
- **Valence electrons of Oxygen (O)**: 6
- **Monovalent atoms (Hydrogen)**: 2 (since there are 2 H atoms)
- **Charge**: 0

Using the formula:
\[ \text{Hybrid orbitals} = \frac{6 + 2 + 0}{2} = \frac{8}{2} = 4 \]
This indicates that H2O has `sp^3` hybridization.

### Conclusion
The pairs of molecules/ions that have central atoms with `sp^2` hybridization are:
- **NO2⁻**
- **BF3**

### Final Answer
The correct option is **NO2⁻ and BF3**.

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