Question

Draw two different Lewis structures for N₂O, one with the arrangement N-O-N, and the other with the arrangement N-N-O. Assign formal charges and circle the more stable structure.

Do either of these structures have a resonance form? Explain.

In addition to measuring bond lengths, how could you distinguish between the N-O-N and N-N-O structures?

Answer 1

The structure I and II are the two different Lewis structures for nitrous oxide with the arrangement of N-O-N and N-N-O.

Determining the formal charge yields the following.

The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge, because it follows the points, the number of atoms with formal charges are minimized and there is no formal charge larger than one. This is again consistent with the preference for having the less electronegative atom in the central position.

This stable structure of nitrous oxide has three resonance structures, out of which one is a major contributor and other one is a minor contributor.

By comparing the structures I and II, the major contributor is II because of the placement of the negative charge. Since oxygen is more electronegative than nitrogen, the negative charge will be stabilized better by oxygen atoms. The structure III will be the minor contributor because it has a greater separation of charge compared with other two. Moreover it shows a positive charge placed on the more electronegative oxygen atom and a very destabilizing -2 charge on nitrogen atom.

By IR spectral studies we can distinguish between N-O-N and N-N-O structures. The experimental value for the N-N stretching vibration is approximately 1595 cm-1 and for N-O stretching vibration at 895 cm-1.