Question

Hybridization of carbon in \[C{H_2} = CH - C{H_2}^ + \] is
(A) \[s{p^2}\]
(B) \[sp\]
(C) \[s{p^3}\]
(D) All of these

Answer 1

Hint: In order to find Hybridization of carbon atom \[C{H_2} = CH - C{H_2}^ + \], we must first know what a hybridisation is. Hybridisation is said to be the intermixing of the two atomic orbitals which have the same energy levels in order to give a hybrid orbital.

Complete step by step answer:
Let us first understand what a hybridisation is. Hybridisation is said to be the intermixing of the two atomic orbitals which have the same energy levels in order to give a hybrid orbital.
Let us see some of the key features of hybridisation:
- The atomic orbitals that have equal energy will be undergoing hybridisation.
- The number of the atomic orbitals which are combined will be equal to the number of hybrid orbitals which are formed.
- Only during the bond formation, the Hybridisation will take place.
- in order to determine the shape of the molecules, we should be aware of the hybridisation.
- In the hybrid orbital, the bigger lobe will have positive signs and the smaller lobe will have negative signs.
Now let us go to the given question:
The structure given is \[C{H_2} = CH - C{H_2}^ + \].
From the above structure, we can see that all the three carbon atoms are attached to three different atoms. Since it is bonded to three different atoms, we can say that the hybridisation of all the three carbon atoms in \[C{H_2} = CH - C{H_2}^ + \] is \[s{p^2}\].
The correct option is option “A” .

Note: We have to remember that different hybridisation present will be having different shapes.

HYBRIDISATION	SHAPE
\[sp\]	Linear
\[s{p^2}\]	Trigonal planar
\[s{p^3}\]	Tetrahedral
\[s{p^3}d\]	Trigonal bipyramidal
\[s{p^3}{d^2}\]	Octahedral