Describe the orbitals that overlap to form the sigma and pi bonds in n2h2
The Correct Answer and Explanation is:
In the molecule N₂H₂ (diazene or diimide), the bonding involves both sigma (σ) and pi (π) bonds. To understand the overlap of orbitals, we need to examine the structure and hybridization of the atoms involved.
Structure and Bonding in N₂H₂:
The structure of N₂H₂ is H–N=N–H, with the nitrogen atoms forming a double bond between them and each nitrogen atom also bonded to one hydrogen atom. The molecule is planar and typically adopts a cis or trans configuration.
Sigma (σ) Bonds:
Sigma bonds are formed by head-on overlap of orbitals.
- N–N sigma bond:
Each nitrogen atom is sp² hybridized. One of the sp² orbitals from each nitrogen overlaps end-to-end to form the N–N sigma bond. - N–H sigma bonds:
The remaining two sp² orbitals on each nitrogen overlap with the 1s orbital of hydrogen atoms to form N–H sigma bonds.
So, in total, there are three sigma bonds in N₂H₂:
- One N–N sigma bond (sp²–sp² overlap)
- Two N–H sigma bonds (sp²–1s overlap)
Pi (π) Bond:
A pi bond is formed by sideways overlap of unhybridized p orbitals.
- Each nitrogen atom has one unhybridized p orbital (since it is sp² hybridized).
- These two p orbitals on adjacent nitrogen atoms overlap side-by-side to form one N=N pi bond.
So, the double bond between the nitrogen atoms consists of:
- One sigma bond (sp²–sp²)
- One pi bond (p–p)
Summary:
- N–N σ bond: sp²–sp² overlap
- N–H σ bonds: sp²–1s overlap
- N=N π bond: unhybridized p–p orbital overlap
This bonding pattern allows the N=N double bond to have both strength and planarity, contributing to the unique geometry and reactivity of diazene.
