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:

• A. $sp^3$, tetrahedral ; $sp^3$, tetrahedral
• B. $sp^2$, trigonal planar ; $sp^3$, tetrahedral
• C. $sp^2$, trigonal planar ; $sp^3$, pyramidal
• D. $sp^3$, tetrahedral ; $sp^2$, trigonal planar

Hint:

Whenever $\text{BF}_3$ accepts a lone pair, boron completes its octet and changes from $sp^2$ trigonal planar to $sp^3$ tetrahedral.

Answer 1

The Correct Option is A

Step 1: Identify the acid and the base.
$\text{BF}_3$ has an empty p orbital on boron, making it a Lewis acid, while $\text{NH}_3$ has a lone pair on nitrogen, making it a Lewis base.
Step 2: Form the adduct.
The nitrogen lone pair fills boron's empty orbital, creating a coordinate bond, \[ \text{F}_3\text{B} \leftarrow \text{NH}_3 \]
Step 3: Count bonds around boron.
Boron now has three B to F bonds plus the new B to N bond, so four sigma bonds and a steric number of 4.
Step 4: Assign boron's hybridization.
A steric number of 4 means $sp^3$ hybridization with tetrahedral geometry around boron.
Step 5: Count bonds around nitrogen.
Nitrogen has three N to H bonds plus the donor bond to boron, again four sigma bonds and steric number 4.
Step 6: Assign nitrogen's hybridization.
So nitrogen is also $sp^3$ with tetrahedral geometry, giving option 1.
\[ \boxed{\text{B: } sp^3 \text{ tetrahedral; N: } sp^3 \text{ tetrahedral}} \]