Question:
$\text{BF}_3$ reacts with $\text{NH}_3$ in 1:1 ratio and gives 'X'. The hybridization and geometry around B and N atoms in 'X' respectively are:
$\text{BF}_3$ reacts with $\text{NH}_3$ in 1:1 ratio and gives 'X'. The hybridization and geometry around B and N atoms in 'X' respectively are:
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Whenever $\text{BF}_3$ accepts a lone pair, boron completes its octet and changes from $sp^2$ trigonal planar to $sp^3$ tetrahedral.
Updated On: Jun 15, 2026
- $sp^3$, tetrahedral ; $sp^3$, tetrahedral
- $sp^2$, trigonal planar ; $sp^3$, tetrahedral
- $sp^2$, trigonal planar ; $sp^3$, pyramidal
- $sp^3$, tetrahedral ; $sp^2$, trigonal planar
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The Correct Option is A
Solution and Explanation
Concept:
$\text{BF}_3$ is a Lewis acid and $\text{NH}_3$ is a Lewis base.
They form a coordinate bond:
\[
\text{F}_3\text{B}\leftarrow\text{NH}_3
\]
Step 1: Hybridization around boron After accepting the lone pair from nitrogen, boron forms four sigma bonds. Therefore, \[ \text{Steric number}=4 \] \[ \Rightarrow sp^3 \text{ hybridization} \] with tetrahedral geometry.
Step 2: Hybridization around nitrogen Nitrogen now also has four sigma bonds. Hence, \[ \text{Steric number}=4 \] \[ \Rightarrow sp^3 \text{ hybridization} \] with tetrahedral geometry. Therefore, \[ \boxed{\text{B: }sp^3,\ \text{tetrahedral;\quad N: }sp^3,\ \text{tetrahedral}} \]
Step 1: Hybridization around boron After accepting the lone pair from nitrogen, boron forms four sigma bonds. Therefore, \[ \text{Steric number}=4 \] \[ \Rightarrow sp^3 \text{ hybridization} \] with tetrahedral geometry.
Step 2: Hybridization around nitrogen Nitrogen now also has four sigma bonds. Hence, \[ \text{Steric number}=4 \] \[ \Rightarrow sp^3 \text{ hybridization} \] with tetrahedral geometry. Therefore, \[ \boxed{\text{B: }sp^3,\ \text{tetrahedral;\quad N: }sp^3,\ \text{tetrahedral}} \]
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