Resonance in X₂Y can be represented as X = X⁺ = Y⁻ ↔ :X ≡ X⁺ − Y: The enthalpy of formation of X₂Y (X(g) + 1/2Y₂(g) → X₂Y(g)) is 80 kJ mol⁻¹. The magnitude of resonance energy of X₂Y is ______ kJ mol⁻¹ (nearest integer value).

 Question:

Resonance in $X_2Y$ can be represented as

$$\mathrm{X}={\mathrm{X}}^{+}={\mathrm{Y}}^{-}\leftrightarrow :\mathrm{X}\equiv {\mathrm{X}}^{+}-\mathrm{Y}:$$

The enthalpy of formation of $X_2Y$ $\big(\mathrm{X(g)} + \tfrac{1}{2}\mathrm{Y_2(g)} \rightarrow X_2Y(g)\big)$is 80 kJ mol⁻¹. The magnitude of resonance energy of $X_2Y$ is _____ kJ mol⁻¹ (nearest integer value).

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Short Solution (Text):

Step 1: Identify the prototype

The resonance set $\mathrm{:X \equiv X^{+} - Y:} \leftrightarrow \mathrm{X = X^{+} = Y^{-}}$is the classic pattern of $\mathrm{N_2O}$ (take $X = \mathrm{N}$, $Y = \mathrm{O}$).

Step 2: Use the given formation enthalpy

For this species, the extra stabilization due to resonance (resonance energy) is taken as the magnitude corresponding to the given energetic stabilization relative to a localized single structure. With $\Delta H_f^\circ$ provided as $80\ \mathrm{kJ\,mol^{-1}}$⁻¹ for formation from 

$\mathrm{X(g)} + \tfrac{1}{2}\mathrm{Y_2(g)}$, the resonance stabilization is taken to be 80 kJ mol⁻¹ (magnitude).

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✅ Final Answer:
Resonance energy (magnitude) = 80 kJ mol⁻¹ (nearest integer)

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Conclusion – Video Solution:

The resonance pattern maps to $\mathrm{N_2O}$. Using the provided enthalpy of formation for the specified atomic route, the magnitude of resonance energy is $80\ \mathrm{kJ\,mol^{-1}}$.