Question

A friend tells you \({{\rm{N}}_{\rm{2}}}\)has three \({\rm{\pi }}\)bonds due to overlap of the three p-orbitals on each N atom. Do you agree?

Short answer

In the given structure; \({N_2},N \equiv N\), there is one sigma and two pi bonds because one sigma bond is formed by the axial overlapping of two p-orbital of each nitrogen atom. After the formation of one sigma bond; remaining p-orbital overlaps by sidewise and will from two pi bond also.

Step-by-step solution

Definition of Concept

The sigma bond is formed when atomic orbitals involved in the formation of a molecular orbital overlap along the internuclear axis.

The bond formed is known as a pi-bond when p-orbitals involved in bond formation overlap sideways.

Explain \({{\rm{N}}_{\rm{2}}}\) has three \({\rm{\pi }}\) bonds due to overlap of the three p-orbitals on each N atom

Considering the given information:

\({N_2},N \equiv N\)

Only a single bond is always a sigma bond, whereas a double bond has one sigma bond and one pi bond, and a triple bond has one sigma bond and two pi bonds.

Therefore, one sigma bond is formed by the axial overlapping of two p-orbitals of each nitrogen atom, the structure \({N_2},N \equiv N\) has one sigma and two pi bonds. Following the formation of one sigma bond, the remaining p-orbital overlaps sideways, resulting in the formation of two pi bonds.