Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2Hmolecules.

Asked by Pragya Singh | 1 year ago |  41

1 Answer

Solution :-

C2H4

Electronic configuration of carbon atom in excited state is given below:

6C: \( 1s^{2}2s^{1}2p_{x}^{1}2p_{y}^{1}2p_{z}^{1}\)

In the formation of C2H4 (ethane) molecule 1 sp2 orbital of C- atom overlaps sp2 orbital of other C- atom. Thus, forming a C-C sigma bond.

The 2 remaining sp2 orbital of every C- atom forms sp2-s \( \sigma\) bond with 2 H- atoms. One c- atom having unhybridized orbital overlaps with the unhybridized orbital of other C- atom and forms a pie bond.

C2H2

In the formation of \( C_2H_2\) molecule, each C-atom is sp hybridized with two 2p-orbitals in an unhybridized state. One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a C-C sigma bond. The second sp orbital if each C-atom overlaps a half-filled 1s-orbital to form a \( \sigma\) bond. The two unhybridized 2p-orbitals of the first carbon undergo sidewise overlap with the 2p orbital of another carbon atom, thereby forming two pi (n) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two n-bonds.

Answered by Abhisek | 1 year ago

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