Which of following is (are) correct for `B` and `N` in `NH_(3).BF_(3)` adduct ? .

Options

A. Both have `sp^(3)` hybrid orbitals
B. Both have tetrahedral structures
C. `N` is `sp^(3)` hybridised while `B` is `sp^(2)` hybridised .
D. `N` in `NH_(3)` is pyramidal, while `B` in `BF_(3)` is planar .

Explanation

To analyze the adduct `NH3.BF3` and determine the correct statements regarding the hybridization and structure of nitrogen (N) and boron (B), we can follow these steps:

### Step 1: Determine the Hybridization of NH3
1. **Count the Valence Electrons of Nitrogen**: Nitrogen has 5 valence electrons.
2. **Bonding with Hydrogen**: Each hydrogen atom requires 1 electron for bonding. Since there are 3 hydrogen atoms, nitrogen will use 3 of its electrons to form bonds with hydrogen.
3. **Lone Pair Calculation**: After forming 3 bonds, nitrogen has 1 electron left, which forms a lone pair.
4. **Hybridization Calculation**: The hybridization is determined by the formula:
\[ \text{Hybridization} = \text{Number of bond pairs} + \text{Number of lone pairs} \]
Here, nitrogen has 3 bond pairs and 1 lone pair:
\[ \text{Hybridization} = 3 + 1 = 4 \quad \Rightarrow \quad \text{sp}^3 \]

### Step 2: Determine the Hybridization of BF3
1. **Count the Valence Electrons of Boron**: Boron has 3 valence electrons.
2. **Bonding with Fluorine**: Each fluorine atom requires 1 electron for bonding. Boron will use all 3 of its electrons to bond with 3 fluorine atoms.
3. **Lone Pair Calculation**: Boron has no electrons left for lone pairs.
4. **Hybridization Calculation**: Boron has 3 bond pairs and 0 lone pairs:
\[ \text{Hybridization} = 3 + 0 = 3 \quad \Rightarrow \quad \text{sp}^2 \]

### Step 3: Forming the Adduct NH3.BF3
1. **Coordinate Bond Formation**: In the adduct, nitrogen donates its lone pair to boron, which has vacant p orbitals. This forms a coordinate bond between nitrogen and boron.
2. **Re-evaluating Hybridization in the Adduct**:
- **For Nitrogen**: After forming the coordinate bond, nitrogen now has 4 bonds (3 with hydrogen and 1 with boron), so it is still sp³ hybridized.
- **For Boron**: After accepting the lone pair from nitrogen, boron also has 4 bonds (3 with fluorine and 1 with nitrogen), so it becomes sp³ hybridized as well.

### Step 4: Determine the Molecular Geometry
1. **Molecular Geometry of NH3**: The molecular geometry of NH3 is trigonal pyramidal due to the lone pair on nitrogen.
2. **Molecular Geometry of BF3**: The molecular geometry of BF3 is trigonal planar.

### Conclusion
- Both nitrogen and boron in the adduct `NH3.BF3` are sp³ hybridized.
- The molecular geometry of nitrogen in NH3 is pyramidal, while boron in BF3 is planar.

### Final Answers
1. Both have sp³ hybrid orbitals: **Correct**
2. Both have tetrahedral structure: **Correct**
3. Nitrogen is sp³ hybridized and boron is sp² hybridized: **Incorrect**
4. Nitrogen in NH3 is pyramidal and boron in BF3 is planar: **Incorrect**