Question

Hybridization of ${CH_3^ - }$ is:
A.$s{p^3}$
B.$s{p^2}$
C.$sp$
D.$ds{p^2}$

Answer 1

Hint:To answer this question, you must recall the concept of hybridization. The concept of mixing of atomic orbitals in order to form new hybrid orbitals that possess different shapes and energies as compared to the original parent atomic orbitals is known as hybridisation.

Complete answer:
Hybrid orbitals are suitable to form chemical bonds of equal energies. Also hybridization of orbitals leads to the formation of more stable compounds because hybrid orbitals have lower energy than the unhybrid orbitals.
We know that carbon has the atomic number 6. The number of valence electrons is 4. The ground state electronic configuration of carbon can be written as
$C:\left[ {He} \right]2{s^2}2{p^2}$
One electron is excited from the $2s$ orbital to $2p$ orbital to increase the covalency. The electronic configuration in this excited state is given as $C:\left[ {He} \right]2{s^1}2{p^3}$
Now we know that, since carbon forms three bonds, it has one free electron. The negative charge signifies another electron, thus resulting in a lone pair. Carbon has three bond pairs and one lone pair. It undergoes a hybridization of $s{p^3}$.
The geometry of the molecule is tetrahedral with three hybridized orbitals forming three bonds and the lone pair as the fourth hybrid orbital. The molecule is pyramidal in shape.

Hence, the correct answer is A.

Note:
To find the geometry of a molecule, we use the VSEPR theory. The Valence shell electron pair repulsion theory proposes that the hybridized orbitals in an atom arrange themselves in such a way so as to minimize the repulsion between them, hence determining the geometry of a molecule on the basis of its hybridization.