Question

What are hybridization states of each carbon atom in the following compounds? $\left( C{{H}_{2}}=C=O \right),\text{ }\left( C{{H}_{3}}CH=C{{H}_{2}} \right),\text{ }\left[ {{\left( C{{H}_{3}} \right)}_{2}}CO \right],\text{ }\left( C{{H}_{2}}=CHCN \right),\text{ }\left( {{C}_{6}}{{H}_{6}} \right).$

Answer 1

We know that the hybridization is said to be the intermixing of the two atomic orbitals which have the same energy levels in order to give a hybrid orbital. The atomic orbitals that have equal energy will be undergoing hybridization. The number of the atomic orbitals which are combined will be equal to the number of hybrid orbitals which are formed.

Complete answer:
The carbon atoms hybridize their outer orbital before forming the bonds as they have only three orbitals to hybridize. So, they use and two orbitals to hybridize to form hybrid orbitals and leave the other electrons unchanged. The three hybrid orbitals arrange themselves at to each other in the plane. Out of the three hybrid orbitals two orbitals overlap axially with orbitals of the neighboring carbon atoms on the both sides to form sigma bonds. In benzene each carbon atom is joined to three other atoms-one hydrogen and two carbons. For finding the hybridization of the carbon atom, the rules are as follows:
If the carbon atom has a single bond, the hybridization is $s{{p}^{3}}.$
If the carbon atom has a double bond, then the hybridization is $s{{p}^{2}}.$
If the carbon atom has a triple bond, then the hybridization is$sp.$
The carbon atom in benzene lacks the number of unpaired electrons to form bonds. The hybridization states of each carbon atom in the following compounds are given below.
$\left( \underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{s{{p}^{{}}}}{\mathop{C}}\,=\underset{{}}{\mathop{O}}\, \right),\text{ }\left( \underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\, \right),\text{ }\left[ \underset{s{{p}^{3}}}{\mathop{{{\left( C{{H}_{3}} \right)}_{2}}}}\,\underset{s{{p}^{2}}}{\mathop{CO}}\, \right],\text{ }\left( \underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,\underset{s{{p}^{3}}}{\mathop{CN}}\, \right),\text{ }\left( \underset{s{{p}^{2}}}{\mathop{{{C}_{6}}{{H}_{6}}}}\, \right).$
The left p orbital overlaps sidewise with similar orbitals of the other carbon atoms on either side to form two sets of pi-bonds. So, the resultant pi-cloud is formed over all the six carbon atoms. This is known as the delocalization of electron charge.

Note:
Remember that all the orbitals in an atom are not of the same energy, so when the atoms attach with other atoms then these energies redistribute among themselves and become equal in energy, and this process is known as hybridization.