Resonance in X_2Y can be represented as shown | vedclass

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(A) The resonance energy is calculated as $\Delta H_{R.E.} = \Delta H_{f(exp)} - \Delta H_{f(theo)}$.Given $\Delta H_{f(exp)} = 80 \ kJ \ mol^{-1}$.Fo

Vedclass · Accepted Answer

$98$

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(A) The resonance energy is calculated as $\Delta H_{R.E.} = \Delta H_{f(exp)} - \Delta H_{f(theo)}$.Given $\Delta H_{f(exp)} = 80 \ kJ \ mol^{-1}$.For the theoretical enthalpy of formation,we consider the reaction: $X \equiv X(g) + \frac{1}{2} Y = Y(g) ightarrow X=X=Y(g)$.$\Delta H_{f(theo)} = (BE_{X \equiv X} + \frac{1}{2} BE_{Y=Y}) - (BE_{X=X} + BE_{X=Y})$.Substituting the given values: $\Delta H_{f(theo)} = (940 + \frac{1}{2} 	imes 500) - (410 + 602)$.$\Delta H_{f(theo)} = (940 + 250) - (1012) = 1190 - 1012 = 178 \ kJ \ mol^{-1}$.Now,$\Delta H_{R.E.} = 80 - 178 = -98 \ kJ \ mol^{-1}$.The magnitude of resonance energy is $|-98| = 98 \ kJ \ mol^{-1}$.

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