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Chapter 9: Problem 64

(a) Write a single Lewis structure for \(\mathrm{N}_{2} \mathrm{O},\) and determine the hybridization of the central \(\mathrm{N}\) atom. (b) Are there other possible Lewis structures for the molecule? (c) Would you expect \(\mathrm{N}_{2} \mathrm{O}\) to exhibit delocalized \(\pi\) bonding?

Short Answer

Expert verified
(a) The single Lewis structure for N2O is O = N = N, with the central N atom having an sp hybridization. (b) There is another possible Lewis structure (N - N ≡ O), but it is less stable due to formal charges, so O = N = N is more favorable. (c) N2O does not exhibit delocalized π bonding as there are no resonance structures involving π bond rearrangement.

Step by step solution

01

Determine the total number of valence electrons

First, we need to count the total number of valence electrons in the N2O molecule. Nitrogen (N) has 5 valence electrons, and Oxygen (O) has 6 valence electrons. So, the total number of valence electrons in N2O is 5 + 5 + 6 = 16.
02

Write a single Lewis structure for N2O

Now, we can draw a single Lewis structure for N2O: 1. Connect the atoms with a single bond. 2. Complete the octets of the outer atoms (Oxygen in this case) by adding lone pairs. 3. Place any remaining electrons on the central atom (Nitrogen in this case). By following these steps, we get the following Lewis structure for N2O: O = N = N
03

Determine the hybridization of the central N atom

To determine the hybridization of an atom, we use the formula: Hybridization = (Number of Bonding Electron Regions + Number of Lone Pairs) In the N2O Lewis structure that we drew, the central N atom has two bonding electron regions and no lone pairs. Therefore, the hybridization of the central N atom is: Hybridization = (2 + 0) = 2 Since the number 2 corresponds to sp hybridization, the hybridization of the central N atom is sp.
04

Discuss possible alternative Lewis structures

There is another possible Lewis structure for N2O: N - N ≡ O However, this structure would result in a less stable molecule due to the positive formal charge on the central N atom and a negative formal charge on the terminal N atom. Therefore, the structure O = N = N is more favorable.
05

Determine if N2O exhibits delocalized π bonding

A molecule exhibits delocalized π bonding when it has multiple resonance structures involving π bond rearrangement. In the case of N2O, there is no resonance due to the lack of alternative stable structures involving π bond rearrangement. In both O = N = N and N - N ≡ O structures, the π bonds do not move between different atoms. Therefore, N2O does not exhibit delocalized π bonding.

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