In which of the following pair both the species have `sp^(3)` hybridization ?

Options

A. `H_(2)`S, `BF_(3)`
B. `SiF_(4), BeH_(2)`
C. `NF_(3), H_(2)O`
D. `NF_(3),BF_(3)`

Answer

To determine which pair of species both have `sp^3` hybridization, we will calculate the hybridization of each species based on their steric numbers. The steric number is calculated as the sum of the number of sigma bonds and the number of lone pairs.

### Step-by-Step Solution:

1. **Identify the Species**: We will analyze the given species one by one to determine their hybridization.

2. **Calculate Hybridization for H2S**:
- **Valence Electrons**: Sulfur (S) is in group 16 and has 6 valence electrons. Each hydrogen (H) contributes 1 electron.
- **Lewis Structure**:
- Sulfur: 6 electrons
- 2 Hydrogens: 2 electrons
- Total: 6 + 2 = 8 electrons
- **Lone Pairs**: Sulfur has 2 lone pairs (6 - 2 from H).
- **Sigma Bonds**: There are 2 sigma bonds (S-H).
- **Steric Number**: 2 (sigma bonds) + 2 (lone pairs) = 4.
- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

3. **Calculate Hybridization for BF3**:
- **Valence Electrons**: Boron (B) is in group 13 and has 3 valence electrons. Each fluorine (F) contributes 1 electron.
- **Lewis Structure**:
- Boron: 3 electrons
- 3 Fluorines: 3 electrons
- Total: 3 + 3 = 6 electrons
- **Lone Pairs**: Boron has no lone pairs.
- **Sigma Bonds**: There are 3 sigma bonds (B-F).
- **Steric Number**: 3 (sigma bonds) + 0 (lone pairs) = 3.
- **Hybridization**: Since the steric number is 3, the hybridization is `sp^2`.

4. **Calculate Hybridization for SiF4**:
- **Valence Electrons**: Silicon (Si) is in group 14 and has 4 valence electrons. Each fluorine (F) contributes 1 electron.
- **Lewis Structure**:
- Silicon: 4 electrons
- 4 Fluorines: 4 electrons
- Total: 4 + 4 = 8 electrons
- **Lone Pairs**: Silicon has no lone pairs.
- **Sigma Bonds**: There are 4 sigma bonds (Si-F).
- **Steric Number**: 4 (sigma bonds) + 0 (lone pairs) = 4.
- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

5. **Calculate Hybridization for BeH2**:
- **Valence Electrons**: Beryllium (Be) is in group 2 and has 2 valence electrons. Each hydrogen (H) contributes 1 electron.
- **Lewis Structure**:
- Beryllium: 2 electrons
- 2 Hydrogens: 2 electrons
- Total: 2 + 2 = 4 electrons
- **Lone Pairs**: Beryllium has no lone pairs.
- **Sigma Bonds**: There are 2 sigma bonds (Be-H).
- **Steric Number**: 2 (sigma bonds) + 0 (lone pairs) = 2.
- **Hybridization**: Since the steric number is 2, the hybridization is `sp`.

6. **Calculate Hybridization for NF3**:
- **Valence Electrons**: Nitrogen (N) is in group 15 and has 5 valence electrons. Each fluorine (F) contributes 1 electron.
- **Lewis Structure**:
- Nitrogen: 5 electrons
- 3 Fluorines: 3 electrons
- Total: 5 + 3 = 8 electrons
- **Lone Pairs**: Nitrogen has 1 lone pair (5 - 3).
- **Sigma Bonds**: There are 3 sigma bonds (N-F).
- **Steric Number**: 3 (sigma bonds) + 1 (lone pair) = 4.
- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

7. **Calculate Hybridization for H2O**:
- **Valence Electrons**: Oxygen (O) is in group 16 and has 6 valence electrons. Each hydrogen (H) contributes 1 electron.
- **Lewis Structure**:
- Oxygen: 6 electrons
- 2 Hydrogens: 2 electrons
- Total: 6 + 2 = 8 electrons
- **Lone Pairs**: Oxygen has 2 lone pairs (6 - 2).
- **Sigma Bonds**: There are 2 sigma bonds (O-H).
- **Steric Number**: 2 (sigma bonds) + 2 (lone pairs) = 4.
- **Hybridization**: Since the steric number is 4, the hybridization is `sp^3`.

### Conclusion:
From the calculations, we find that:
- H2S: `sp^3`
- BF3: `sp^2`
- SiF4: `sp^3`
- BeH2: `sp`
- NF3: `sp^3`
- H2O: `sp^3`

The pair that has both species with `sp^3` hybridization is **H2S and NF3** or **SiF4 and H2O**. However, the question specifically asks for a pair, and based on the analysis, **H2S and H2O** or **NF3 and H2O** can be considered as valid pairs.

### Final Answer:
The correct pair where both species have `sp^3` hybridization is **H2S and H2O** or **NF3 and H2O**.