Question

What type of hybridization of each carbon atom in the following compounds?
(a) \[C{H_3}Cl\]

Answer 1

Hint: In chemical science, orbitally hybridizing (or hybridizing) means mixing atomic orbits into new hybrid orbits (with energy, forms, etc. different from those of the atomic orbits) suitable for the combination of electrons in the theory of valence bond chemistry.

Complete answer:
Sigma bonds = number + lone pair. = number of Sigma bonds \[C{H_3}Cl\] = 3 sigma links from \[C\]&\[H\] to 1 from \[C\] to \[Cl\].
No lone pair is present, because carbon has four electrons of valence and all of them are connected (3 with hydrogen and 1 with chlorine \[Cl\]).
Now, hybridization = \[(3 + 1) = \,4\]
\[s{p^3}\] (1 s & 3 p).
so \[s{p^3}\]hybridization is the answer to the given question.
As we know the hybrid orbitals are useful in explaining molecular geometry and atomic bonding properties because they are symmetrically arranged in space, and they are symmetrically disposed in space Hybrid orbitals are typically formed by combining atomic orbitals with similar energies.
\[s{p^3}\]
The term "hybridization" refers to the bonding of atoms as seen from the perspective of an atom. To form a bond with the four hydrogen atoms in a tetrahedrally coordinated carbon (for example, methane (\[C{H_4}\]), the carbon must have four orbitals that have the correct symmetry.

Note:
The lowest energy is obtained if the four bonds are equivalent, which means that they must be formed from equivalent orbitals on the carbon atoms. The four\[s{p^3}\] hybrids are a set of four equivalent orbitals that are linear combinations of the valence-shell (core orbitals are almost never involved in bonding) s and p wave functions.