Given that: (i) $$\Delta_f H°$$ of $$N_2O$$ is 82 kJ mol$$^{-1}$$ (ii) Bond energies of $$N \equiv N$$, $$N = N$$, $$O = O$$ and $$N = O$$ are 946, 418, 498 and 607 kJ mol$$^{-1}$$ respectively. The resonance energy of $$N_2O$$ is :

To find the resonance energy of $$ \text{N}_2\text{O} $$, we start by recalling that resonance energy is the difference between the actual enthalpy of formation and the calculated enthalpy of formation assuming no resonance. The actual $$ \Delta_f H^\circ $$ for $$ \text{N}_2\text{O} $$ is given as 82 kJ mol⁻¹.

Next, we calculate the enthalpy of formation using bond energies without considering resonance. The formation reaction for $$ \text{N}_2\text{O} $$ is:

$$ \text{N}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{N}_2\text{O}(\text{g}) $$

For this reaction, we need to break bonds in the reactants and form bonds in the product. The reactants are $$ \text{N}_2 $$ and $$ \text{O}_2 $$. In $$ \text{N}_2 $$, there is one $$ \text{N} \equiv \text{N} $$ bond with bond energy 946 kJ mol⁻¹. In $$ \frac{1}{2} \text{O}_2 $$, there is half of an $$ \text{O} = \text{O} $$ bond, so the energy is $$ \frac{1}{2} \times 498 = 249 $$ kJ mol⁻¹. The total energy required to break bonds in the reactants is:

$$ 946 + 249 = 1195 \text{ kJ mol}^{-1} $$

Now, for the product $$ \text{N}_2\text{O} $$, we assume a structure without resonance. The common hypothetical structure is $$ ^{-}\text{N} = \text{N}^{+} = \text{O} $$, which has one $$ \text{N} = \text{N} $$ bond and one $$ \text{N} = \text{O} $$ bond. The bond energies are given as 418 kJ mol⁻¹ for $$ \text{N} = \text{N} $$ and 607 kJ mol⁻¹ for $$ \text{N} = \text{O} $$. The total energy released when these bonds are formed is:

$$ 418 + 607 = 1025 \text{ kJ mol}^{-1} $$

The enthalpy change for the reaction, which is the calculated $$ \Delta_f H^\circ $$ without resonance, is the energy absorbed minus the energy released:

$$ \Delta_f H^\circ (\text{calculated}) = 1195 - 1025 = 170 \text{ kJ mol}^{-1} $$

The resonance energy is defined as the actual $$ \Delta_f H^\circ $$ minus the calculated $$ \Delta_f H^\circ $$:

$$ \text{Resonance energy} = 82 - 170 = -88 \text{ kJ mol}^{-1} $$

The negative value indicates stabilization due to resonance. Comparing with the options, -88 kJ corresponds to option A.

Hence, the correct answer is Option A.