In which of the following the central atom does not use `sp^3` hybrid orbitals in its bonding

Options

A. `BeF_(3) ^(-)`
B. `OH_(3) ^(+)`
C. `NH_(2) ^(-)`
D. `NH_(3)`

Answer

To determine which of the given options has a central atom that does not use `sp^3` hybrid orbitals in its bonding, we will calculate the steric number for each option using the formula:

**Steric Number (SN) = 1/2 * (V + M + Charge)**

Where:
- V = number of valence electrons of the central atom
- M = number of monovalent atoms attached to the central atom
- Charge = add the charge if it's an anion and subtract if it's a cation.

Let's analyze each option step by step:

### Step 1: Analyze BeF3^(-)
1. **Identify the central atom**: Beryllium (Be)
2. **Valence electrons (V)**: Be has 2 valence electrons.
3. **Monovalent atoms (M)**: There are 3 fluorine (F) atoms attached.
4. **Charge**: The charge is -1 (anion), so we add 1.

**Calculation**:
\[ \text{SN} = \frac{1}{2} \times (2 + 3 + 1) = \frac{1}{2} \times 6 = 3 \]

- **Hybridization**: Since the steric number is 3, the hybridization is `sp^2`.

### Step 2: Analyze OH3^(+)
1. **Identify the central atom**: Oxygen (O)
2. **Valence electrons (V)**: O has 6 valence electrons.
3. **Monovalent atoms (M)**: There are 3 hydrogen (H) atoms attached.
4. **Charge**: The charge is +1 (cation), so we subtract 1.

**Calculation**:
\[ \text{SN} = \frac{1}{2} \times (6 + 3 - 1) = \frac{1}{2} \times 8 = 4 \]

- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

### Step 3: Analyze NH2^(-)
1. **Identify the central atom**: Nitrogen (N)
2. **Valence electrons (V)**: N has 5 valence electrons.
3. **Monovalent atoms (M)**: There are 2 hydrogen (H) atoms attached.
4. **Charge**: The charge is -1 (anion), so we add 1.

**Calculation**:
\[ \text{SN} = \frac{1}{2} \times (5 + 2 + 1) = \frac{1}{2} \times 8 = 4 \]

- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

### Step 4: Analyze Ammonia (NH3)
1. **Identify the central atom**: Nitrogen (N)
2. **Valence electrons (V)**: N has 5 valence electrons.
3. **Monovalent atoms (M)**: There are 3 hydrogen (H) atoms attached.
4. **Charge**: There is no charge.

**Calculation**:
\[ \text{SN} = \frac{1}{2} \times (5 + 3 + 0) = \frac{1}{2} \times 8 = 4 \]

- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

### Conclusion:
The only option where the central atom does not use `sp^3` hybrid orbitals is **BeF3^(-)**, which is `sp^2` hybridized.

### Final Answer:
**The central atom does not use `sp^3` hybrid orbitals in BeF3^(-).**

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