Question

Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in \(C_2H_4\) and \(C_2H_2\) molecules.

Answer 1

\(C_2H_4\) : 
The electronic configuration of \(C-atom\) in the excited state is:
\(_6C\)=\(1s^2 2s^1 2p_x^1 2p_y^1 2p_z^1\)

In the formation of an ethane molecule \((C_2H_4)\), one \(sp ^2\) hybrid orbital of carbon overlaps a \(sp ^2\) hybridized orbital of another carbon atom, thereby forming a \(C-C\) sigma bond.
The remaining two \(sp ^2\) orbitals of each carbon atom form a \(sp ^2\) \(-s\) sigma bond with two hydrogen atoms. The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak \(\pi-bond\).

\(C_2H_2\): 
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 of 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 \(pi (\pi)\) bonds between carbon atoms. 
Hence, the triple bond between two carbon atoms is made up of one sigma and two \(\pi-bonds\).