Question

The hybridisation and geometry of $B$ and $N$ in $ [H_3B {

• A. $sp^3$ , tetrahedral and $sp^3$ pyramidal
• B. $sp^3$ , pyramidal and $sp^3$ tetrahedral
• C. $sp^3$ , pyramidal and $sp^3$ pyramidal
• D. $sp^3$ , tetrahedral and $sp^3$ tetrahedral

Answer 1

Answer: (C)

$sp^3$ , pyramidal and $sp^3$ pyramidal

Answer 2

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 $B$. 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.