Question

Which one of the following is a planar molecule ?

• A. $N{H}_3$
• B. $H_3{O}^{+}$
• C. $BC{l}_3$
• D. $PC{l}_3$

Answer 1

Answer: (C)

$BC{l}_3$

$H=\frac{1}{2}[V+M-C+A]$
where,
H= Number of orbitals involved in hybridization.
V=Valence electrons of central atom.
M- Number of monovalent atoms linked to central atom.
C= Charge of cation.
A= Charge of anion.
Now consider the options
A) $N{H}_3$
$H=\frac{1}{2}[5+3]=4$.
$\Rightarrow s{p}^3$ hybridized state .
Trigonal Pyramid in shape.
B) $H_3{O}^{+}$
$H=\frac{1}{2}[6+3-1]=4$.
$\Rightarrow s{p}^3$ hybridized state .
Tetrahedral in shape.
C) $BC{l}_3$
$H=\frac{1}{2}[3+3]=3$.
$\Rightarrow s{p}^2$ hybridized state .
Trigonal Planar in shape.
D) $PC{l}_3$
$H=\frac{1}{2}[5+3]=4$.
$\Rightarrow s{p}^3$ hybridized state .
Trigonal Pyramid in shape.
Hence option C is the right answer.