Question

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

• A. N: Tetrahedral, ; B: tetrahedral,
• B. N: pyramidal,; B: pyramidal,
• C. N: pyramidal, ; B: planar,
• D. N: pyramidal, ; B: tetrahedral,

Answer 1

The correct option is A

N: Tetrahedral, ; B: tetrahedral,

We have already seen the shapes and geometries of both $B{F}_3$ and $N{H}_3$. Let us look at them once again:

(For B)ep = $\frac{V+L+A-C}{2}$ = $\frac{3+3}{2}$ = 3 $\Rightarrow$ Trigonal Planer (No lone pairs)

For N $\Rightarrow$ ep = $\frac{V+L+A-C}{2}$ = $\frac{5+3}{2}$ $\Rightarrow$ Tetrahedral(with one lone pair)

$B{F}_3$ $\Rightarrow$

Three ${sp}^2$ hybrid orbitals which will overlap with Flourine

Four ${sp}^3$ hybrid orbitals which will overlap with Hydrogen

Therefore , $B{F}_3$ $\to$ Trigonal Planar , ${sp}^2$

$N{H}_3$ $\to$ Tetrahedral,${sp}^3$

Now, Let us try and use the VSEPR theory to calculate the shape and geometry of the 1:1 complex of $B{F}_3$ and $N{H}_3$. Once the lone pair on $N$ atom in $N{H}_3$ forms a coordinate covalent bond with the $B{F}_3$ molecule, there is a positive charge on the $N$ atom and a negative charge on the $B$ atom.

(For B)ep = $\frac{V+L+A-C}{2}$ = $\frac{3+4+1-0}{2}$ = 4 $\Rightarrow$ Tetrahedral

For N $\Rightarrow$ ep = $\frac{V+L+A-C}{2}$ = $\frac{5+4+0-1}{2}$ $\Rightarrow$ Tetrahedral