Draw all of the resonance structures for the N_(2)O_(4) molecule and determine the average N-O bond order. The skeletal structure of the molecule is able[[O,O],[N,N],[O,]] 8.113 Draw all of the resonance structures for the N,O4 mol ecule and determine the average N-O bond order. The skeletal structure of the molecule is 0 0 N N 0 0
The Correct Answer and Explanation is:
✅ Correct Answer: The molecule N₂O₄ has four major resonance structures. In each, the N–N bond remains a single bond, while the N–O bonds alternate between single and double bonds across the resonance forms. The average N–O bond order is 1.5.
🧠 Explanation
Dinitrogen tetroxide (N₂O₄) is a covalent molecule composed of two nitrogen atoms and four oxygen atoms. Its skeletal structure is symmetric, with each nitrogen atom bonded to two oxygen atoms and to the other nitrogen atom. The key to understanding its bonding lies in resonance theory, which describes how electrons can be delocalized across multiple atoms, leading to multiple valid Lewis structures.
In N₂O₄, the N–N bond is always a single bond in all resonance forms. The variation occurs in the N–O bonds. Each nitrogen is bonded to two oxygen atoms, and the double bonds can shift among these oxygens. This results in four major resonance structures, each with two N=O double bonds and two N–O single bonds. The double bonds are not fixed in one location but are delocalized across the molecule.
To determine the average N–O bond order, we consider the total number of N–O bonds and how many are double versus single across the resonance forms. In each resonance structure, there are four N–O bonds: two are single (bond order 1), and two are double (bond order 2). The total bond order contribution is:
(2 × 1) + (2 × 2) = 2 + 4 = 6
Dividing this total by the number of N–O bonds (4) gives the average bond order:
6 ÷ 4 = 1.5
This means that each N–O bond in N₂O₄ has partial double bond character, consistent with experimental observations such as bond lengths that are intermediate between typical N–O single and double bonds. The resonance hybrid is more stable than any individual structure and reflects this delocalization.
Thus, the average N–O bond order in N₂O₄ is 1.5.
