Question

Specify the co-ordination geometry around and hybridization of $N$ and $B$-atoms in a $1:1$ complex of ${BF}_3$ and ${NH}_3$ :

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

Answer 1

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

No doubt $N{H}_3$ and $B{F}_3$ have $s{p}^3$ (pyramidal) and $s{p}^2$ (coplanar) hybridization, respectively, having one lone pair of electron on N atom which is responsible for pyramidal shape of $N{H}_3$ inspite of $s{p}^3$ hybridization. However, as soon as it is coordinated to $B{F}_3$, both attain tetrahedral geometry and acquire $s{p}^3$ hybridization .