Question

In benzene, each carbon atom undergoes:
(A) sp hybridization
(B) \[s{{p}^{2}}\]hybridization
(C) \[s{{p}^{3}}\]hybridization
(D) d\[s{{p}^{2}}\]hybridization

Answer 1

Hint: Benzene is the organic compound having the molecular formula hexacarbon hexahydride. It is the ring shaped compound attached to each carbon atom by planar rings. It is also called a hydrocarbon because it has only hydrogen and carbon atoms. Benzene is a compound which is also used for the production of polystyrene. In benzene there are formation of triple bonds which are weak due to which it acquires electrophilic character and undergoes reactions with neutrophiles.

Complete step by step solution:
In benzene each carbon atom is attached to two other carbon atoms, while in non ring structures the one carbon is attached to one carbon atom. So in total each carbon is attached to two carbon atoms and one hydrogen atom. So they form \[s{{p}^{2}}\] hybridization for connecting the atoms by planar rings. The carbon atom forms sigma bonds with other three atoms that are two carbon atoms and one hydrogen atom. So the geometry of each carbon atom in benzene is trigonal planar geometry. In this hybridization it is must to achieve the bond angle of 120 degree and this angle is achieved in benzene. Benzene forms a regular hexagon. It is a stable compound because the lengths of carbon- carbon bonds are somewhere equal either it be the single bond or be the double bond. There is the presence of delocalised electrons above and below the planar ring due to which the benzene becomes stable. In benzene each carbon atom has a p- orbital due to which it can participate in p- bonding.

Hence, (B) option is correct.

Note: students should know what is hybridization and geometry of each organic compound. They should know the meaning of \[s{{p}^{2}}\],\[s{{p}^{3}}\],and sp hybridization. You should give examples of each hybridization in definition to gain marks. The hybridised orbitals of benzene are not parallel to each other but they overlap laterally to each other for the formation of pi bonds which is not the part of pi electrons.