Question

Hybridization of $C{H_3}^ - $ is:
A) $s{p^3}$
B) $s{p^2}$
C) $sp$
D) $ds{p^2}$

Answer 1

Hint:To solve this question, we must first understand the concept of hybridization. Then we need to assess and follow the correct steps and rules to find the Hybridization of the required compound and then only we can conclude the correct answer.


Complete solution:Before we move forward with the solution of this given question, let us first understand some basic concepts 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. 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.
The atomic orbitals of similar energy are mixed together such as the mixing of two $'s'$ orbitals or two $'p'$ orbitals or mixing of an $'s'$ orbital with a $'p'$ orbital or $'s'$ orbital with a $'d'$ orbital.
Types of Hybridization:
$sp$ hybridization: is observed when one s and one p orbital in the same main shell of an atom mix to form two new equivalent orbitals.
$s{p^2}$ hybridization: is observed when one s and two p orbitals of the same shell of an atom mix to form 3 equivalent orbits.
$s{p^3}$ hybridization: When one ‘s’ orbital and 3 ‘p’ orbitals belonging to the same shell of an atom mix together to form four new equivalent orbits.
$s{p^3}d$ hybridization: involves the mixing of 3p orbitals and 1d orbital to form 5 $s{p^3}d$ hybridized orbitals of equal energy.
$s{p^3}{d^2}$ hybridization: has $1s$ , $3p$ and $2d$ orbitals, that undergo intermixing to form 6 identical $s{p^3}{d^2}$ hybrid orbitals.
Now, we will go through the given compound:
The carbanion has $3$ bonding pairs and one lone pair. Thus VSEPR theory predicts a tetrahedral geometry and a trigonal planar structure.
A tetrahedral electron geometry corresponds to $s{p^3}$ hybridisation.

Thus, the correct option is (A).

Note: The bigger lobe of the hybrid orbital always has a positive sign, while the smaller lobe on the opposite side has a negative sign.