Question

In \[{{C}}{{{H}}_3}^ - \] hybridization on carbon in carbanion is:
A. ${{s}}{{{p}}^3}$
B. ${{s}}{{{p}}^3}{{{d}}^2}$
C. ${{s}}{{{p}}^2}$
D. ${{s}}{{{p}}^3}{{d}}$

Answer 1

Hint: First let’s know what hybridization is. When atomic orbitals having different energies are combined, it forms new orbitals, i.e. molecular orbitals. These orbitals have the same energies and shape. The hybridization of a molecule can be determined with the help of VSEPR theory.

Complete step by step answer:
In \[{{C}}{{{H}}_3}^ - \], carbon is the central atom. From the negative charge in the chemical formula of a molecule, it is obvious that the given molecule is an anion. Anion is formed when the molecule is gained by an electron. When a methyl group gains an electron, it becomes \[{{C}}{{{H}}_3}^ - \]. Carbon has four valence electrons in its outermost shell. Carbon atom is bonded to three hydrogen atoms and there is an extra electron in it. As we know that the methane molecule is ${{s}}{{{p}}^3}$ hybridized and it has tetrahedral geometry. But in \[{{C}}{{{H}}_3}^ - \], this tetrahedral geometry gets distorted.
When we use steric numbers, we can also determine the hybridization of the \[{{C}}{{{H}}_3}^ - \] molecule. From the chemical formula, carbon has three bonded pairs which are connected with hydrogen atoms and one lone pair of electrons. Thus it is ${{s}}{{{p}}^3}$ hybridized.
Thus the \[{{C}}{{{H}}_3}^ - \] anion is ${{s}}{{{p}}^3}$ hybridized.

So, the correct answer is Option A.

Additional information:
This method which we have used for finding the hybridization of carbon, can be used for finding the hybridization of other elements. Hybridized molecules are more stable than unhybridized molecules.

Note: Similar to \[{{C}}{{{H}}_3}^ - \] molecule, ${{C}}{{{H}}_3}^ + $ is formed by losing an electron from methyl group. But the hybridization is different from \[{{C}}{{{H}}_3}^ - \] ion. Thus the electrons are similar to one s-orbital and two p-orbital.