Which of the following pairs have same hybridisation ?

Options

A. `CH_(3)^(-)` and `CH_(3)^(+)`
B. `NH_(3)` and `SO_(2)`
C. `SO_(4)^(2-)` and `BF_(4)^(-)`
D. `BF_(3)` and `NF_(3)`

Answer

To determine which pairs of compounds have the same hybridization, we will calculate the hybridization for each compound using the formula provided in the transcript. The formula for calculating the number of hybrid orbitals (x) is:

\[ x = \frac{1}{2} \left( \text{number of valence electrons on the central atom} + \text{number of monovalent atoms} - \text{charge on cation} + \text{charge on anion} \right) \]

### Step-by-Step Solution:

1. **Option 1: CH3- and CH3+**
- **For CH3-:**
- Central atom: Carbon (C) has 4 valence electrons.
- Monovalent atoms: 3 Hydrogens (H).
- Charge: -1 (since it is an anion).
- Calculation:
\[ x = \frac{1}{2} \left( 4 + 3 - 0 + 1 \right) = \frac{1}{2} \times 8 = 4 \quad \Rightarrow \text{Hybridization: } sp^3 \]
- **For CH3+:**
- Central atom: Carbon (C) has 4 valence electrons.
- Monovalent atoms: 3 Hydrogens (H).
- Charge: +1 (since it is a cation).
- Calculation:
\[ x = \frac{1}{2} \left( 4 + 3 - 1 + 0 \right) = \frac{1}{2} \times 6 = 3 \quad \Rightarrow \text{Hybridization: } sp^2 \]
- **Conclusion:** CH3- is sp³ and CH3+ is sp². **Not the same hybridization.**

2. **Option 2: NH3 and SO2**
- **For NH3:**
- Central atom: Nitrogen (N) has 5 valence electrons.
- Monovalent atoms: 3 Hydrogens (H).
- Charge: 0 (neutral molecule).
- Calculation:
\[ x = \frac{1}{2} \left( 5 + 3 - 0 + 0 \right) = \frac{1}{2} \times 8 = 4 \quad \Rightarrow \text{Hybridization: } sp^3 \]
- **For SO2:**
- Central atom: Sulfur (S) has 6 valence electrons.
- Monovalent atoms: 2 Oxygens (O) (each oxygen is bivalent).
- Charge: 0 (neutral molecule).
- Calculation:
\[ x = \frac{1}{2} \left( 6 + 0 - 0 + 0 \right) = \frac{1}{2} \times 6 = 3 \quad \Rightarrow \text{Hybridization: } sp^2 \]
- **Conclusion:** NH3 is sp³ and SO2 is sp². **Not the same hybridization.**

3. **Option 3: SO4²- and BF4-**
- **For SO4²-:**
- Central atom: Sulfur (S) has 6 valence electrons.
- Monovalent atoms: 0 (no monovalent atoms).
- Charge: -2 (since it is an anion).
- Calculation:
\[ x = \frac{1}{2} \left( 6 + 0 - 0 + 2 \right) = \frac{1}{2} \times 8 = 4 \quad \Rightarrow \text{Hybridization: } sp^3 \]
- **For BF4-:**
- Central atom: Boron (B) has 3 valence electrons.
- Monovalent atoms: 4 Fluorines (F).
- Charge: -1 (since it is an anion).
- Calculation:
\[ x = \frac{1}{2} \left( 3 + 4 - 0 + 1 \right) = \frac{1}{2} \times 8 = 4 \quad \Rightarrow \text{Hybridization: } sp^3 \]
- **Conclusion:** Both SO4²- and BF4- are sp³. **Same hybridization.**

4. **Option 4: BF3 and NF3**
- **For BF3:**
- Central atom: Boron (B) has 3 valence electrons.
- Monovalent atoms: 3 Fluorines (F).
- Charge: 0 (neutral molecule).
- Calculation:
\[ x = \frac{1}{2} \left( 3 + 3 - 0 + 0 \right) = \frac{1}{2} \times 6 = 3 \quad \Rightarrow \text{Hybridization: } sp^2 \]
- **For NF3:**
- Central atom: Nitrogen (N) has 5 valence electrons.
- Monovalent atoms: 3 Fluorines (F).
- Charge: 0 (neutral molecule).
- Calculation:
\[ x = \frac{1}{2} \left( 5 + 3 - 0 + 0 \right) = \frac{1}{2} \times 8 = 4 \quad \Rightarrow \text{Hybridization: } sp^3 \]
- **Conclusion:** BF3 is sp² and NF3 is sp³. **Not the same hybridization.**

### Final Answer:
The pair that has the same hybridization is **Option 3: SO4²- and BF4-**.