In which of the following the central atom does not use `sp^(2)` hybrid orbitals in its bonding

Options

A. `CH_(3)^(+)`
B. `overset(-)CH_(3)`
C. `NF_(3)`
D. `BeF_(3)^(-)`

Explanation

To determine which of the given options has a central atom that does not use `sp²` hybrid orbitals in its bonding, we will calculate the hybridization of each molecule based on the steric number. The steric number (S) can be calculated using the formula:

\[ S = \frac{1}{2} \left( \text{Valence electrons of central atom} + \text{Monovalent atoms attached} - \text{Cationic charge} + \text{Anionic charge} \right) \]

Let's analyze each option step by step.

### Step 1: Analyze the first option (CS₃⁺)

1. **Identify the central atom**: Carbon (C)
2. **Valence electrons of carbon**: 4
3. **Monovalent atoms attached**: 3 (3 hydrogen atoms)
4. **Cationic charge**: 1 (since it is CS₃⁺)
5. **Anionic charge**: 0

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \left( 4 + 3 - 1 + 0 \right) = \frac{1}{2} \left( 6 \right) = 3 \]

- **Hybridization for S = 3**: `sp²`

### Step 2: Analyze the second option (C₃H₃⁻)

1. **Identify the central atom**: Carbon (C)
2. **Valence electrons of carbon**: 4
3. **Monovalent atoms attached**: 3 (3 hydrogen atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 1 (since it is C₃H₃⁻)

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \left( 4 + 3 - 0 + 1 \right) = \frac{1}{2} \left( 8 \right) = 4 \]

- **Hybridization for S = 4**: `sp³`

### Step 3: Analyze the third option (NF₃)

1. **Identify the central atom**: Nitrogen (N)
2. **Valence electrons of nitrogen**: 5
3. **Monovalent atoms attached**: 3 (3 fluorine atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 0

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \left( 5 + 3 - 0 + 0 \right) = \frac{1}{2} \left( 8 \right) = 4 \]

- **Hybridization for S = 4**: `sp³`

### Step 4: Analyze the fourth option (BeF₃⁻)

1. **Identify the central atom**: Beryllium (Be)
2. **Valence electrons of beryllium**: 2
3. **Monovalent atoms attached**: 3 (3 fluorine atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 1 (since it is BeF₃⁻)

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \left( 2 + 3 - 0 + 1 \right) = \frac{1}{2} \left( 6 \right) = 3 \]

- **Hybridization for S = 3**: `sp²`

### Conclusion

From our analysis:
- **CS₃⁺**: sp²
- **C₃H₃⁻**: sp³
- **NF₃**: sp³
- **BeF₃⁻**: sp²

The central atom does not use `sp²` hybrid orbitals in **C₃H₃⁻** and **NF₃**. Therefore, the correct options are **C₃H₃⁻** and **NF₃**.