The reaction between ${\mathrm{BF}}_3$ and ${\mathrm{NH}}_3$ forms an adduct (a Lewis acid-base complex) where ${\mathrm{NH}}_3$ acts as a Lewis base and ${\mathrm{BF}}_3$ acts as a Lewis acid.

Step 1: Analyze ${\mathrm{NH}}_3$

In ${\mathrm{NH}}_3$, the nitrogen atom has 3 bond pairs (with H atoms) and 1 lone pair. The steric number is $3+1=4$. Therefore, the hybridization of N is ${\mathrm{sp}}^3$. Due to the presence of one lone pair, the geometry is pyramidal, not tetrahedral. However, in the context of coordination geometry around the central atom, if we consider the electron domains, it's tetrahedral electron geometry, but molecular geometry is pyramidal. When it forms a complex, the lone pair forms a coordinate bond, effectively becoming a bond pair.

Step 2: Analyze ${\mathrm{BF}}_3$

In ${\mathrm{BF}}_3$, the boron atom has 3 bond pairs (with F atoms) and no lone pairs. The steric number is 3. Therefore, the hybridization of B is ${\mathrm{sp}}^2$, and its geometry is trigonal planar.

Step 3: Analyze the complex formation

When ${\mathrm{NH}}_3$ donates its lone pair to ${\mathrm{BF}}_3$, a coordinate covalent bond is formed between N and B. The complex formed is ${\mathrm{H}}_3\mathrm{N}\to {\mathrm{BF}}_3$.

• For Nitrogen (N) in the complex: After donating its lone pair to form a coordinate bond, N is now bonded to three H atoms and one B atom. It has 4 bond pairs and 0 lone pairs. The steric number is 4. Thus, the hybridization of N becomes ${\mathrm{sp}}^3$, and its coordination geometry becomes tetrahedral.
• For Boron (B) in the complex: After accepting the lone pair from N, B is now bonded to three F atoms and one N atom. It has 4 bond pairs and 0 lone pairs. The steric number changes from 3 to 4. Thus, the hybridization of B changes from ${\mathrm{sp}}^2$ to ${\mathrm{sp}}^3$, and its coordination geometry becomes tetrahedral.

Step 4: Evaluate the options

• A) $\mathrm{N}$ : tetrahedral, ${\mathrm{sp}}^3;\mathrm{B}$ : tetrahedral, ${\mathrm{sp}}^3$ - This matches our analysis.
• B) $\mathrm{N}$ : pyramidal, ${\mathrm{sp}}^3;\mathrm{B}$ : pyramidal, ${\mathrm{sp}}^3$ - N becomes tetrahedral in the complex, not pyramidal. B becomes tetrahedral, not pyramidal.
• C) $\mathrm{N}$ : pyramidal, ${\mathrm{sp}}^3;\mathrm{B}$ : planar, ${\mathrm{sp}}^2$ - N becomes tetrahedral, and B becomes tetrahedral (${\mathrm{sp}}^3$).
• D) $\mathrm{N}$ : pyramidal, ${\mathrm{sp}}^3;\mathrm{B}$ : tetrahedral, ${\mathrm{sp}}^3$ - N becomes tetrahedral in the complex, not pyramidal.

Correct Answer: (A)