Specify the coordination geometry around and the hybridisation of `N` and `B` atoms in `1 : 1` complex of `BF_(3)` and `NH_(3)`.

Options

A. N: tetrahedral `sp^(3)`, B: tetrahedral `sp^(3)`
B. N: pyramidal, `sp^(3)`, B : pyramidal `sp^(2)`
C. N: pyramidal, `sp^(3)` , B : planar `sp^(2)`
D. N: pyramidal, `sp^(3)` , B : tetrahedral, `sp^(3)` .

Explanation

To determine the coordination geometry and hybridization of the nitrogen (N) and boron (B) atoms in the 1:1 complex of BF₃ and NH₃, we can follow these steps:

### Step 1: Analyze the Hybridization of BF₃
- Boron trifluoride (BF₃) has three bond pairs and no lone pairs.
- The hybridization of boron in BF₃ is **sp²**.
- The coordination geometry around boron is **trigonal planar**.

### Step 2: Analyze the Hybridization of NH₃
- Ammonia (NH₃) has three bond pairs (N-H bonds) and one lone pair.
- The hybridization of nitrogen in NH₃ is **sp³**.
- The coordination geometry around nitrogen is **tetrahedral**.

### Step 3: Formation of the 1:1 Complex
- When BF₃ reacts with NH₃, a coordinate bond is formed where the lone pair of nitrogen is donated to boron.
- This results in the formation of the complex BF₃·NH₃.

### Step 4: Determine the Hybridization in the Complex
- In the complex, the nitrogen atom still has three N-H bonds and now forms a coordinate bond with BF₃, maintaining a total of four bonding interactions.
- Therefore, nitrogen in the complex remains **sp³** hybridized with a tetrahedral geometry.

- For boron, when it forms a coordinate bond with NH₃, it effectively has four bonding interactions (three F atoms and one N atom). Hence, boron also becomes **sp³** hybridized, leading to a tetrahedral geometry.

### Conclusion
- In the 1:1 complex of BF₃ and NH₃:
- The nitrogen atom (N) is **sp³ hybridized** with a **tetrahedral geometry**.
- The boron atom (B) is also **sp³ hybridized** with a **tetrahedral geometry**.