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

Hint: 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.