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  4. The hybridisation and geometry of B and N in [H3B arrow NH3] are, respectively

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Q. The hybridisation and geometry of $ B $ and $ N $ in $ [H_3B \leftarrow NH_3] $ are, respectively

AMUAMU 2016Chemical Bonding and Molecular Structure

Solution:

Since, $BF_3$ is electron deficient and needs two electrons to complete its octet. It acts as a Lewis acid and reacts with $NH$, which is a Lewis base due to the presence of lone pair of $e$ - on $N$-atom. Acceptance of a pair of electrons from $NH_3$ completes the octet of $B$.
image
Ceptral atom: Boron
Hybridisation: $sp^2$
Geometry: tetrahedral
In $BF_3 \to NH_3$. Both $N$ and $B$ contains atoms with
$sp^3$ -hybridisation with tetrahedral geometry.