Which of the following species contains three bond pairs and two ione pair around the central atom ?

Options

A. `NH_(2)^(-)`
B. `ClF_(3)`
C. `H_(2)`O
D. `BF_(3)`

Answer

To determine which species contains three bond pairs and two lone pairs around the central atom, we will analyze each option provided in the question step-by-step.

### Step 1: Identify the central atom and its valence electrons
For each species, we need to identify the central atom and count its valence electrons.

### Step 2: Use the formula to find the number of hybrid orbitals
The formula to find the hybridization is:
\[ \text{Hybridization} = \frac{\text{Valence electrons of central atom} + \text{Number of monovalent atoms} + \text{Charge}}{2} \]
Where:
- Valence electrons are the electrons in the outermost shell of the central atom.
- Monovalent atoms are those that form one bond (like H, F, etc.).
- Charge is added if negative and subtracted if positive.

### Step 3: Calculate for each species

1. **NH2⁻ (Amide ion)**
- Central atom: Nitrogen (N)
- Valence electrons: 5 (N)
- Monovalent atoms: 2 (H)
- Charge: -1 (add 1)
- Calculation:
\[ \text{Hybridization} = \frac{5 + 2 + 1}{2} = \frac{8}{2} = 4 \quad (\text{sp}^3) \]
- Bond pairs: 2 (from H atoms)
- Lone pairs: 2 (remaining electrons)
- **Does not match (needs 3 bond pairs).**

2. **ClF3 (Chlorine trifluoride)**
- Central atom: Chlorine (Cl)
- Valence electrons: 7 (Cl)
- Monovalent atoms: 3 (F)
- Charge: 0
- Calculation:
\[ \text{Hybridization} = \frac{7 + 3 + 0}{2} = \frac{10}{2} = 5 \quad (\text{sp}^3d) \]
- Bond pairs: 3 (from F atoms)
- Lone pairs: 2 (remaining electrons)
- **Matches (3 bond pairs and 2 lone pairs).**

3. **H2O (Water)**
- Central atom: Oxygen (O)
- Valence electrons: 6 (O)
- Monovalent atoms: 2 (H)
- Charge: 0
- Calculation:
\[ \text{Hybridization} = \frac{6 + 2 + 0}{2} = \frac{8}{2} = 4 \quad (\text{sp}^3) \]
- Bond pairs: 2 (from H atoms)
- Lone pairs: 2 (remaining electrons)
- **Does not match (needs 3 bond pairs).**

4. **BF3 (Boron trifluoride)**
- Central atom: Boron (B)
- Valence electrons: 3 (B)
- Monovalent atoms: 3 (F)
- Charge: 0
- Calculation:
\[ \text{Hybridization} = \frac{3 + 3 + 0}{2} = \frac{6}{2} = 3 \quad (\text{sp}^2) \]
- Bond pairs: 3 (from F atoms)
- Lone pairs: 0 (no remaining electrons)
- **Does not match (needs 2 lone pairs).**

### Conclusion
The species that contains three bond pairs and two lone pairs around the central atom is **ClF3**.