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 `sp³` hybridization, we will analyze each pair given in the question step by step.

### Step 1: Understand the Hybridization Formula
The formula for determining hybridization is:
\[ \text{Hybridization} = \frac{\text{Valence electrons of central atom} + \text{Number of monovalent atoms} + \text{Charge}}{2} \]

### Step 2: Analyze Each Pair

#### Pair 1: H₂S and BF₃
1. **H₂S (Hydrogen Sulfide)**
- Central atom: Sulfur (S)
- Valence electrons: 6 (Group 16)
- Monovalent atoms: 2 (Hydrogens)
- Calculation:
\[ \text{Hybridization} = \frac{6 + 2}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad \text{sp}^3 \]

2. **BF₃ (Boron Trifluoride)**
- Central atom: Boron (B)
- Valence electrons: 3 (Group 13)
- Monovalent atoms: 3 (Fluorines)
- Calculation:
\[ \text{Hybridization} = \frac{3 + 3}{2} = \frac{6}{2} = 3 \quad \Rightarrow \quad \text{sp}^2 \]

**Conclusion for Pair 1:** Not both have `sp³` hybridization.

#### Pair 2: SiF₄ and BeH₂
1. **SiF₄ (Silicon Tetrafluoride)**
- Central atom: Silicon (Si)
- Valence electrons: 4 (Group 14)
- Monovalent atoms: 4 (Fluorines)
- Calculation:
\[ \text{Hybridization} = \frac{4 + 4}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad \text{sp}^3 \]

2. **BeH₂ (Beryllium Hydride)**
- Central atom: Beryllium (Be)
- Valence electrons: 2 (Group 2)
- Monovalent atoms: 2 (Hydrogens)
- Calculation:
\[ \text{Hybridization} = \frac{2 + 2}{2} = \frac{4}{2} = 2 \quad \Rightarrow \quad \text{sp} \]

**Conclusion for Pair 2:** Not both have `sp³` hybridization.

#### Pair 3: NF₃ and H₂O
1. **NF₃ (Nitrogen Trifluoride)**
- Central atom: Nitrogen (N)
- Valence electrons: 5 (Group 15)
- Monovalent atoms: 3 (Fluorines)
- Calculation:
\[ \text{Hybridization} = \frac{5 + 3}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad \text{sp}^3 \]

2. **H₂O (Water)**
- Central atom: Oxygen (O)
- Valence electrons: 6 (Group 16)
- Monovalent atoms: 2 (Hydrogens)
- Calculation:
\[ \text{Hybridization} = \frac{6 + 2}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad \text{sp}^3 \]

**Conclusion for Pair 3:** Both have `sp³` hybridization.

#### Pair 4: NF₃ and BF₃
1. **NF₃ (Nitrogen Trifluoride)** (already calculated)
- Hybridization: `sp³`

2. **BF₃ (Boron Trifluoride)** (already calculated)
- Hybridization: `sp²`

**Conclusion for Pair 4:** Not both have `sp³` hybridization.

### Final Answer
The pair in which both species have `sp³` hybridization is **Pair 3: NF₃ and H₂O**.

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