Question

In which molecule, all atoms are coplanar?
A.${\text{C}}{{\text{H}}_{\text{4}}}$
B. ${\text{B}}{{\text{F}}_{\text{3}}}$
C.${\text{P}}{{\text{F}}_{\text{3}}}$
D.${\text{N}}{{\text{H}}_{\text{3}}}$

Answer 1

To answer this question,recall the formula to calculate the hybridization of a molecule. The value of the hybridisation number will help determine the shape which is dependent on the hybridization number.
Formula used:
For neutral molecules: ${\text{Hybridisation = lone pair + bond pair}}$

Complete step by step answer:
The important thing to remember is that the molecules exhibiting $s{p^2}$hybridisation are planar. This is because hybridisation leads to triangular planar shape. $s{p^2}$is observed when one s and two p orbitals of the same shell of an atom mix to form 3 equivalent orbitals.
First, determine the hybridization of the given molecules and then determine the shape and geometry of the molecules:
${\text{C}}{{\text{H}}_{\text{4}}}$​−The central atom forms 4 bonds and has 0 lone pair. The hybridisation is $s{p^3}$ with tetrahedral geometry. This shape is non-planar, the option is incorrect and is eliminated.
${\text{B}}{{\text{F}}_{\text{3}}}$ − The central atom forms 3 bonds and has 0 lone pair. The hybridisation is $s{p^2}$ with trigonal planar geometry. This shape is planar, the option is correct.
${\text{P}}{{\text{F}}_{\text{3}}}$- The central atom forms 3 bonds and has 1 lone pair. The hybridisation is $s{p^3}$ with trigonal pyramidal geometry. This shape is non-planar, the option is incorrect and is eliminated
${\text{N}}{{\text{H}}_{\text{3}}}$ − The central atom forms 3 bonds and has 1 lone pair. The hybridisation is $s{p^3}$ with trigonal pyramidal geometry. This shape is non-planar, the option is incorrect and is eliminated.
So, ${\text{B}}{{\text{F}}_{\text{3}}}$ the molecule is coplanar and is the correct answer.

Hence, the correct option is B.

Note:
Even if you are not able to calculate the hybridisation using the above-mentioned you can find the hybridization $(X)$ using the formula: $\frac{1}{2}({\text{V}} + {\text{H}} - {\text{C}} + {\text{A}})$ where
${\text{V}}$= Number of valence electrons in the central atom
${\text{H}}$= Number of surrounding monovalent atoms
${\text{C}}$= Cationic charge
${\text{A}}$= Anionic charge. The value of X will determine the hybridisation of the molecule.
If $X$=2 then $sp$; =3 then $s{p^2}$ ; =4 then $s{p^3}$; =5 then $s{p^3}d$ ; =6 then $s{p^3}{d^2}$ ; =7 then $s{p^3}{d^3}$ hybridization.