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

Answer

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

### Step 1: Understand the Nature of BF₃ and NH₃
- BF₃ (Boron Trifluoride) is an electron-deficient molecule, meaning it has less than an octet of electrons around the boron atom. It acts as a Lewis acid and can accept a pair of electrons.
- NH₃ (Ammonia) has a lone pair of electrons on the nitrogen atom and acts as a Lewis base, donating its lone pair to BF₃.

**Hint:** Identify the roles of BF₃ and NH₃ in the reaction based on their electron configurations.

### Step 2: Formation of the Complex
- When BF₃ reacts with NH₃, a coordinate bond is formed. The nitrogen atom donates its lone pair to the boron atom, resulting in the formation of a complex.

**Hint:** Remember that a coordinate bond involves the donation of a lone pair from one atom to another.

### Step 3: Determine the Hybridization of Boron (B)
- In BF₃, boron is bonded to three fluorine atoms, which means it has three sigma bonds. The hybridization of boron can be determined as follows:
- For three sigma bonds, the hybridization is sp².
- The geometry associated with sp² hybridization is planar.

**Hint:** Count the number of sigma bonds to determine the hybridization of boron.

### Step 4: Determine the Hybridization of Nitrogen (N)
- In NH₃, nitrogen has three bond pairs (from the N-H bonds) and one lone pair. The hybridization can be determined as follows:
- For three bond pairs and one lone pair, the hybridization is sp³.
- The geometry associated with sp³ hybridization is tetrahedral, but due to the presence of a lone pair, the actual geometry is pyramidal.

**Hint:** Consider the presence of lone pairs when determining the geometry of molecules.

### Step 5: Analyze the Complex
- In the BF₃-NH₃ complex, boron now has four sigma bonds (three with fluorine and one with nitrogen). Therefore, the hybridization of boron in the complex changes to sp³, resulting in a tetrahedral geometry.
- Nitrogen also forms four sigma bonds (three with hydrogen and one with boron), maintaining its sp³ hybridization and tetrahedral geometry in the complex.

**Hint:** Re-evaluate the hybridization and geometry after the formation of the complex.

### Conclusion
- The coordination geometry around nitrogen (N) in the complex is tetrahedral with sp³ hybridization.
- The coordination geometry around boron (B) in the complex is also tetrahedral with sp³ hybridization.

### Final Answer
- N is tetrahedral and sp³; B is tetrahedral and sp³.