Given that:  
(i)  ${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\circ }\text{ of }{\mathrm{N}}_2\mathrm{O}\text{ is }82\mathrm{kJ}{\mathrm{mol}}^{-1}$

(ii) Bond energies of $\mathrm{N}\equiv \mathrm{N},\mathrm{N}=\mathrm{N},\mathrm{O}=\mathrm{O}\text{ and }\mathrm{N}=\mathrm{O}$ are 

 946, 418, 498 and 607 kJ mol^-1 respectively, The resonance energy of N₂O is :

${\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{N}}_2\mathrm{O}\left(\mathrm{g}\right)$
$\mathrm{N}\equiv \mathrm{N}\left(\mathrm{g}\right)+\frac{1}{2}(\mathrm{O}=\mathrm{O})\to \mathrm{N}=\mathrm{N}=\mathrm{O}\left(\mathrm{g}\right)$
${{\mathrm{\Delta H}}_{\mathrm{f}}}^{\mathrm{o}}$ = [Energy required for breaking of bonds] —[Energy released for forming of bonds] 

          = [Enthalpies of reactants]-[Enthalpies of products]  
          $$\begin{gathered} =\left({\mathrm{\Delta H}}_{\mathrm{N}\equiv \mathrm{N}}+\frac{1}{2}{\mathrm{\Delta H}}_{\mathrm{U}=\mathrm{O}}-\left({\mathrm{\Delta H}}_{\mathrm{N}=\mathrm{N}}+{\mathrm{\Delta H}}_{\mathrm{N}=\mathrm{O}}\right)\right. \\ =\left(946+\frac{1}{2}\times 498\right)-(418+607)=170\mathrm{kJ}{\mathrm{mol}}^{-1} \end{gathered}$$ 
Resonance energy = 170-82 = 88kJ mol^-1