Question

Hybridization states of C in $CH_3^ +$ and $CH_3^ -$ are:
a.) $s{p^2},s{p^3}$
b.) $s{p^3},s{p^2}$
c.) $sp,s{p^2}$
d.) $s{p^2},sp$

Answer 1

Hint: To find the hybridization of any chemical compounds we will first define the term hybridization and then we have few procedures of steps to be followed which will help us to find the same of the given organic compound in this question.
The term hybridization is defined as the mixing of the atomic orbitals that belong to the same atom but have slightly different energies so that energy is redistributed between them to form the new orbitals with equal energies and the same shape.

Complete answer:
In both $CH_3^ +$ and $CH_3^ -$ carbon is the central atom. The hybridization of carbon therefore involves the hybridization of the two molecules. We have four valence electrons for a carbon atom. The positive charge in molecule $CH_3^ +$ means that an electron is removed. Thus 3 electrons of valence remain in the outermost carbon shell. The electronic configuration of C in $CH_3^ +$ is $1{s^2}2{s^1}2{p^4}$ . These electrons are equivalent to one orbital "s" and two p "orbitals. Thus, one 's' and two 'p' carbon orbitals combine the overlapping of three individual H atoms with the 's' orbitals, resulting in a $s{p^2}$ hybridization.
Similarly in $CH_3^ - ,$ there are three H-and extra-electron-bindings to the carbon atom. Consequently, the tetrahedral geometry with $s{p^3}$ hybridization forms distorted.
Therefore the hybridization of $CH_3^ +$ is $s{p^2}$ and the hybridization of $CH_3^ -$ is $s{p^3}$

Hence, the correct option is A.

Note: The above process we followed to find the hybridization of the carbon atoms can be followed for finding the hybridization of the other elements in the same manner. Hybridization is also greatly enhanced by hybrid orbitals, as their separate, unhybridized counterparts are less energy efficient. As a result, hybridization leads to more stable compounds.