Question

Specify the coordination geometry around and hybridization of Nitrogen and Boron atom in a $1:1$ complex of $B{F_3}$ and $N{H_3}$

Answer 1

Hint: We have to configure the coordinate geometry of the Boron and Nitrogen atom in both cases. Nitrogen is known to be considered in the group $15$ in periodic table and has five valence electrons. Boron is known to be present in group $13$ and it has three valence electrons.

Complete step-by-step answer:
In this complex, the ratio is $1:1$. $N{H_3}$has $3$ bonds present between Nitrogen and the atom Hydrogen but as there is a lone pair present in Nitrogen thus the hybridization of it is $s{p_3}$ and geometry is pyramidal. Compound $B{F_3}$ has $3$ bond pairs between atom Boron and atom Fluoride but $0$ lone pairs thus have the hybridization $s{p_2}$ and the geometry as coplanar. When the complex is formed between the two, there is a bond formed in which the$N{H_3}$donates its lone pair to$B{F_3}$ .It is possible because $B{F_3}$ has one empty ${p_z}$ orbital and $s{p_2}$ hybridization. Thus the Nitrogen atom will have the geometry with $s{p_3}$ hybridization and Boron will also have the geometry with $s{p_3}$ hybridization. The structure is Tetrahedral. The name of the complex is ammonia-borontriflouride complex.

Thus the answer will be Nitrogen: Tetrahedral, $s{p_3}$; Boron: tetrahedral, $s{p_3}$

Note:There are four bonds formed by Nitrogen and Boron in the complex and this is the reason why the hybridization of both Nitrogen and Boron becomes $s{p_3}$. $N{H_3}$ is known as the molecule in which the central atom has both shared as well as unshared pairs of electrons.