Question

Specify the coordination geometry around and hybridization of N and B atoms in a 1:1 complex of $B{F}_{3}andN{H}_3$:

• A. N : tetrahedral , $s{p}^3$ ; B : tetrahedral , $s{p}^3$
• B. N : pyramidal , $s{p}^3$ ;B : pyramidal , $s{p}^3$
• C. N : pyramidal , $s{p}^3$ ;B : planar , $s{p}^2$
• D. N : pyramidal , $s{p}^3$ ;B : tetrahedral , $s{p}^3$

Answer 1

The correct option is A N : tetrahedral , $s{p}^3$ ; B : tetrahedral , $s{p}^3$

Vacant orbital of Boron atom accepts lone pair of nitrogen of $N{H}_3$. The $s{p}^2$ hybrid orbital of $B{F}_3\text{became}s{p}^3$ and Geometry changes to Tetrahedral from trigonal planar, $N$ in $N{H}_3$ $s{p}^3$ hybridise, Geometry is Tetrahedral.