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Resonance Energy Calculation Using Enthalpy Data

Learn how to calculate resonance energy using enthalpy of formation and resonance structures. This method helps solve JEE Chemistry bonding problems..

 Question:

Resonance in X2YX_2Y can be represented as

X=X+=Y        :XX+Y:

The enthalpy of formation of X2YX_2Y (X(g)+12Y2(g)X2Y(g))\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 X2YX_2Y is _____ kJ mol⁻¹ (nearest integer value).

đź“· Question Image:

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).


Short Solution (Text):

Step 1: Identify the prototype

The resonance set :XX+Y:X=X+=Y\mathrm{:X \equiv X^{+} - Y:} \leftrightarrow \mathrm{X = X^{+} = Y^{-}}is the classic pattern of N2O\mathrm{N_2O} (take X=NX = \mathrm{N}, Y=OY = \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 ΔHf\Delta H_f^\circ provided as
80\ \mathrm{kJ\,mol^{-1}}
⁻¹ for formation from 

X(g)+12Y2(g)\mathrm{X(g)} + \tfrac{1}{2}\mathrm{Y_2(g)}, the resonance stabilization is taken to be 80 kJ mol⁻¹ (magnitude).


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


đź“· Solution Image:


Conclusion – Video Solution:

The resonance pattern maps to N2O\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}}
.

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