Question

Which of the following species have a trigonal planar shape?
(A) $C{H_3}^ -$
(B) $C{H_3}^ +$
(C) $B{F_4}^ -$
(D) $Si{H_4}$

Answer 1

To answer this question, you should recall the concept of hybridization. Hybridization is defined as the concept of mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals. This intermixing is based on quantum mechanics. The atomic orbitals of the same energy level can only take part in hybridization and both full-filled and half-filled orbitals can also take part in this process, provided they have equal energy.

Complete step by step solution:
The important thing to remember that the molecules exhibiting $s{p^2}$ hybridisation are planar because this 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.
In the ions: $C{H_3}^ -$ , $B{F_4}^ -$ and $Si{H_4}$ , the central atom is $s{p^3}$ hybridized. Thus, the geometry of these molecules is tetrahedral.
In the methyl carbocation, the central atom i.e. carbon has three bond pairs which result in $s{p^2}$ hybridization.
The structure can be represented as:

The methyl carbocation $(C{H_3}^ + )$ has $s{p^2}$ hybridisation due to which it has a trigonal planar shape. Maintaining a positive charge is easier than negative charge and negative charge makes the shape to pyramidal. The positive charge does not affect the shape and it is trigonal planar.
Therefore, we can conclude that the correct answer to this question is option B.

Note:
Hybridisation of a molecule can be found out using $(X)$ the formula: $\dfrac{1}{2}(V + H - C + A)$ where
$V$ = Number of valence electrons in the central atom
$H$ = Number of surrounding monovalent atoms
$C$ = Cationic charge
$A$ = Anionic charge. The value of X will determine the hybridisation of the molecule. If $X$ is 2 then $sp$ ; is 3 then $s{p^2}$ ; is 4 then $s{p^3}$ ; is 5 then $s{p^3}d$ ; is 6 then $s{p^3}{d^2}$ ; is 7 then $s{p^3}{d^3}$ hybridization.