Question

The${\text{N}}_{\text{2}}\text{O}$molecule is linear and polar.

1. On the basis of this experimental evidence, which arrangement, NNO or$\text{NON}$, is correct? Explain your answer.
2. On the basis of your answer in part a, write the Lewis structure of${\text{N}}_{\text{2}}\text{O}$(including resonance forms). Give the formal charge on each atom and the hybridization of the central atom.
3. How would the multiple bonding in

$\mathit{N}\equiv \mathit{N}-\stackrel{..}{\mathit{O}}\text{:}$

be described in terms of orbitals?

Short answer

1. The $\text{NNO}$structure is correct.
2. Because the terminal nitrogen has one sigma bond and one loan pair of electrons, and the intermediate nitrogen has two sigma bonds, the hybridization of both nitrogen atoms is$\mathrm{Sp}$in${\text{N}}_{\text{2}}\text{O}$ . Because it has one sigma bond and three lone pairs of electrons, the oxygen atom possesseshybridization.
3. It contain $one\sigma$bond$two\pi$ and bonds make up the .

Step-by-step solution

Concept Introduction

Hybridization is the process of mixing orbitals with somewhat varying energies in order to redistribute their energy and create a new set of orbitals with comparable shapes and energies.

Draw the Lewis structure

a)

Nitrous oxide is a chemical molecule that has the formula ${\text{N}}_{\text{2}}\text{O}$and is a nitrogen oxide.

For the given molecular formula, we can draw two Lewis structures.

Calculate the total amount of valence electrons in each atom before drawing the Lewis structure in such a way that each atom has its own octet configuration.

The figure below depicts Lewis constructions.

Now, writing about polarity,

Polarity is the separation of electric charge that causes an electric dipole moment in a molecule or its chemical groups, with a negatively charged end and a positively charged end.

A difference in electronegativity between the connected atoms necessitates the presence of polar bonds in polar molecules.

The NON structure is nonpolar and symmetrical.

Because of the difference in electronegativity between nitrogen and oxygen, we have a polar N-O bond in the NNO structure. We can conclude that $\text{NNO}$is the proper structure because it has been experimentally shown that ${\text{N}}_{\text{2}}\text{O}$is a polar molecule.

Therefore, the answer is the localid="1663845117278" $\text{NNO}$structure is correct.

Draw the resonance structure 

b)

The expected structure for$N_2$is $\text{NNO}$, which is explained in the Solution for Ex. 107a.

The image below depicts the resonance structures for${\text{N}}_{\text{2}}\text{O}$.

The formal charges for each atom are also represented in the diagram above.

The following simple equation is used to calculate formal charges.

$\text{Formal charge = (number of valence electrons) - (electrons in lone pairs + 1 / 2 the number of bonding electrons)}$

Because the terminal nitrogen has one sigma bond and one loan pair of electrons, and the intermediate nitrogen has two sigma bonds, the hybridization of both nitrogen atoms is $\mathrm{Sp}$in ${\text{N}}_{\text{2}}\text{O}$. Because it has one sigma bond and three lone pairs of electrons, the oxygen atom possesses hybridization.

Therefore, the hybridization of nitrogen atoms is $\mathrm{Sp}$in ${\text{N}}_{\text{2}}\text{O}$because it has one sigma bond and three lone pairs of electrons, the oxygen atom possesses ${\text{sp}}^{\text{3}}$hybridization.

Hybridization of compound

c)

The center nitrogen atom has been $\text{s p}$hybridized, establishing two bonds with the oxygen and nitrogen atoms around it. Each nitrogen atom has two p orbitals that overlap and produce two $\pi$bonds. Two $\pi$bonds are perpendicular to each other. This indicates that one $\pi$bond is positioned above and below the line between two nuclei, whereas the other $pi$ bond is placed on the left or right side of the line. As a result, one $\mathrm{s}\mathrm{p}$hybrid orbital produces a $\sigma$bond, and two p orbitals form two $\pi$bonds in multiple bonding.

Hence, bond$\sigma$ and $two\pi$bonds make up the .