Among the coimpounds, `BF_(3), NH_(3), H_(2)O,SF_(4)and BeCl_(2)`, identify the ones in which the central atom has the same type of hybridization.

Options

A. `BF_(3)and NCl_(3)`
B. `NH_(3)and H_(2)O`
C. `BF_(3), NCl _(3) and H_(2)O`
D. `SF_(4)and BeCl_(2)`

Answer

To determine the hybridization of the central atom in the compounds BF₃, NH₃, H₂O, SF₄, and BeCl₂, we will analyze each compound step by step.

### Step 1: Analyze BF₃ (Boron Trifluoride)
- **Valence Electrons**: Boron (B) has 3 valence electrons, and each Fluorine (F) has 7 valence electrons.
- **Bonding**: Boron forms 3 bonds with 3 Fluorine atoms, using all its valence electrons.
- **Lone Pairs**: There are 0 lone pairs on Boron.
- **Steric Number**: 3 (3 bond pairs + 0 lone pairs).
- **Hybridization**: The hybridization is **sp²**.

### Step 2: Analyze NH₃ (Ammonia)
- **Valence Electrons**: Nitrogen (N) has 5 valence electrons, and each Hydrogen (H) has 1 valence electron.
- **Bonding**: Nitrogen forms 3 bonds with 3 Hydrogen atoms, using 3 of its valence electrons.
- **Lone Pairs**: There is 1 lone pair on Nitrogen.
- **Steric Number**: 4 (3 bond pairs + 1 lone pair).
- **Hybridization**: The hybridization is **sp³**.

### Step 3: Analyze H₂O (Water)
- **Valence Electrons**: Oxygen (O) has 6 valence electrons, and each Hydrogen (H) has 1 valence electron.
- **Bonding**: Oxygen forms 2 bonds with 2 Hydrogen atoms, using 2 of its valence electrons.
- **Lone Pairs**: There are 2 lone pairs on Oxygen.
- **Steric Number**: 4 (2 bond pairs + 2 lone pairs).
- **Hybridization**: The hybridization is **sp³**.

### Step 4: Analyze SF₄ (Sulfur Tetrafluoride)
- **Valence Electrons**: Sulfur (S) has 6 valence electrons, and each Fluorine (F) has 7 valence electrons.
- **Bonding**: Sulfur forms 4 bonds with 4 Fluorine atoms, using 4 of its valence electrons.
- **Lone Pairs**: There is 1 lone pair on Sulfur.
- **Steric Number**: 5 (4 bond pairs + 1 lone pair).
- **Hybridization**: The hybridization is **sp³d**.

### Step 5: Analyze BeCl₂ (Beryllium Chloride)
- **Valence Electrons**: Beryllium (Be) has 2 valence electrons, and each Chlorine (Cl) has 7 valence electrons.
- **Bonding**: Beryllium forms 2 bonds with 2 Chlorine atoms, using both of its valence electrons.
- **Lone Pairs**: There are 0 lone pairs on Beryllium.
- **Steric Number**: 2 (2 bond pairs + 0 lone pairs).
- **Hybridization**: The hybridization is **sp**.

### Summary of Hybridizations:
- BF₃: sp²
- NH₃: sp³
- H₂O: sp³
- SF₄: sp³d
- BeCl₂: sp

### Conclusion:
The compounds NH₃ and H₂O have the same type of hybridization, which is **sp³**.