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)`

Explanation

To determine which species in the given pairs has `sp^3` hybridization, we can use the steric number method. The steric number is calculated by adding the number of sigma bonds, lone pairs of electrons, and the magnitude of any negative charges on the atom. If the steric number equals 4, the species is `sp^3` hybridized.

### Step-by-step Solution:

1. **Identify the Species**: We need to analyze the hybridization of the species in the given pairs.

2. **Option A: H2S**
- **Valence Electrons**: Sulfur (S) has 6 valence electrons.
- **Bonding**: H2S has 2 hydrogen atoms bonded to sulfur.
- **Lone Pairs**: Sulfur has 2 lone pairs.
- **Steric Number Calculation**:
- Sigma bonds: 2 (from H2)
- Lone pairs: 2
- Total = 2 + 2 = 4
- **Hybridization**: Since the steric number is 4, H2S is `sp^3` hybridized.

3. **Option B: SiF4**
- **Valence Electrons**: Silicon (Si) has 4 valence electrons.
- **Bonding**: SiF4 has 4 fluorine atoms bonded to silicon.
- **Lone Pairs**: There are no lone pairs on silicon.
- **Steric Number Calculation**:
- Sigma bonds: 4 (from Si-F)
- Lone pairs: 0
- Total = 4 + 0 = 4
- **Hybridization**: Since the steric number is 4, SiF4 is `sp^3` hybridized.

4. **Option C: NF3 and H2O**
- **For NF3**:
- **Valence Electrons**: Nitrogen (N) has 5 valence electrons.
- **Bonding**: NF3 has 3 fluorine atoms bonded to nitrogen.
- **Lone Pairs**: There is 1 lone pair on nitrogen.
- **Steric Number Calculation**:
- Sigma bonds: 3 (from N-F)
- Lone pairs: 1
- Total = 3 + 1 = 4
- **Hybridization**: Since the steric number is 4, NF3 is `sp^3` hybridized.

- **For H2O**:
- **Valence Electrons**: Oxygen (O) has 6 valence electrons.
- **Bonding**: H2O has 2 hydrogen atoms bonded to oxygen.
- **Lone Pairs**: There are 2 lone pairs on oxygen.
- **Steric Number Calculation**:
- Sigma bonds: 2 (from H2)
- Lone pairs: 2
- Total = 2 + 2 = 4
- **Hybridization**: Since the steric number is 4, H2O is `sp^3` hybridized.

5. **Conclusion**:
- From the analysis, both H2S and SiF4 have `sp^3` hybridization, as do NF3 and H2O. Therefore, the correct answer is that both species in option C (NF3 and H2O) have `sp^3` hybridization.

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