91Ó°ÊÓ

Nitrogen has five electrons in the valence shell and oxygen has six electrons in the valence shell. So, the total number of valence electrons in ${\text{N}}_{\text{2}}\text{O}$is sixteen. All the three resonance structures of ${\text{N}}_{\text{2}}\text{O}$ have sixteen valence electrons and full octets. All three resonance structures contain two sigma bonds and two pi bonds. The difference between the three structures is due to the placement of the negative charge. Since the oxygen atom is more electronegative, the negative charge is better stable on the oxygen atom rather than the nitrogen atom. The third structure is the least stable because there is more charge separation than the other two structures and also positive charge present on more electronegative oxygen atoms and -2 formal charge on the nitrogen atom is very unstable.

In structure A, the Formal charge on Nitrogen (a) is +1.

on Nitrogen (b) is -1.

on Oxygen is 0

In structure B, the Formal charge on Nitrogen (a) is 0.

on Nitrogen (b) is +1.

on Oxygen is -1.

In structure C, the Formal charge on Nitrogen (a) is -2.

on Nitrogen (b)is +1.

on Oxygenis +1.

Since oxygen is a more electronegative atom than nitrogen, the positive charge is better stabilized on the nitrogen atom than on the oxygen atom. Due to the positive charge on Oxygen, this resonance structure is not stable.

Nitrous oxide contains$\text{36 - 37\%}$of available oxygen while air contains$\text{21\%}$of oxygen.

Nitrous oxide supports combustion more vigorously than air because it has more oxygen content than air.

The reaction that occurs for combustion is,

$\mathbf{2}{\mathbf{N}}_{\mathbf{2}}\mathbf{O}→\mathbf{2}{\mathbf{N}}_{\mathbf{2}}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}$