Question

Define hybridization.

Answer 1

Hint: The hybridization of methane having formula $C{{H}_{4}}$ is $s{{p}^{3}}$ because the carbon atom is attached with four hydrogen atoms. The hybridization of ethene having formula $C{{H}_{2}}=C{{H}_{2}}$ is $s{{p}^{2}}$ because one each carbon atom is attached with three atoms.

Complete answer:
Each atom has a different configuration so all the elements have different atomic orbitals. When different elements combine to form the orbitals that have the same energy and identical shapes, there will be a formation of a compound. These orbitals are known as hybrid orbitals. So, we can define the process of mixing of the atomic orbitals belonging to the same atom but having slightly different energies, to form new orbitals of equal energies and identical shape after the distribution of energy takes place is known as hybridization.
There are some points about hybridization:
(i)- Only the orbitals that have approximate equal energies and belong to the same atom or ion can undergo hybridization.
(ii)- The number of formed hybrid orbitals is equal to the number of mixing atomic orbitals.
(iii)- The process of hybridization only takes place during bond formation, and it does not take place in the isolated atom.
(iv)- The geometry of the molecule can be predicted by the hybridization of the molecule.
There are many types of hybridization like $\text{sp},\text{ s}{{\text{p}}^{2}},\text{ s}{{\text{p}}^{3}},\text{ ds}{{\text{p}}^{3}},\text{ }{{\text{d}}^{2}}\text{s}{{\text{p}}^{3}},\text{ s}{{\text{p}}^{3}}\text{d, s}{{\text{p}}^{3}}{{\text{d}}^{2}}$. The $\text{sp},\text{ s}{{\text{p}}^{2}},\text{ s}{{\text{p}}^{3}}$ hybridizations are mostly used in organic compounds. The $\text{ds}{{\text{p}}^{2}},\text{ }{{\text{d}}^{2}}\text{s}{{\text{p}}^{3}},\text{ s}{{\text{p}}^{3}}\text{d, s}{{\text{p}}^{3}}{{\text{d}}^{2}}$hybridization are mostly used in coordination compounds.

Note:
In $sp$ hybridization, one s-orbital and one p-orbital are involved in the bond formation, in $s{{p}^{2}}$ hybridization, one s-orbital and two p-orbitals are involved in the bond formation, in $s{{p}^{3}}{{d}^{2}}$ hybridization, one s-orbital, three p-orbitals, and two d-orbitals are involved in the formation of a bond, etc.