JEE Chemistry 2025 Question with Solution

Given: The enthalpy of formation of $X_2Y$ is $80 \, \text{kJ mol}^{-1}$. The resonance forms indicate delocalization and hence extra stabilization.

Find: The magnitude of the resonance energy of $X_2Y$.

From the bond-energy calculation shown in the solution:

$$\text{Energy required to break bonds} = 940 + \frac{1}{2}\times 500 = 1190 \, \text{kJ mol}^{-1}$$ $$\text{Energy released on bond formation} = 410 + 602 = 1012 \, \text{kJ mol}^{-1}$$

Therefore, the expected enthalpy of formation without resonance is

$$\Delta H_{\text{expected}} = 1190 - 1012 = 178 \, \text{kJ mol}^{-1}$$

The actual enthalpy of formation is

$$\Delta H_{\text{actual}} = 80 \, \text{kJ mol}^{-1}$$

So, the resonance energy is

$$\text{RE} = \left|\Delta H_{\text{expected}} - \Delta H_{\text{actual}}\right| = |178 - 80| = 98 \, \text{kJ mol}^{-1}$$

Therefore, the magnitude of resonance energy is $98 \, \text{kJ mol}^{-1}$.

Given:

• Break $X=X$ bond: $940 \, \text{kJ mol}^{-1}$
• Break $\frac{1}{2}Y=Y$ bond: $250 \, \text{kJ mol}^{-1}$
• Form one $X-X$ bond: $410 \, \text{kJ mol}^{-1}$
• Form one $X-Y$ bond: $602 \, \text{kJ mol}^{-1}$
• Actual enthalpy of formation of $X_2Y$: $80 \, \text{kJ mol}^{-1}$

Find: Resonance energy.

First calculate total bond breaking energy:

$$940 + 250 = 1190 \, \text{kJ mol}^{-1}$$

Now calculate total bond formation energy:

$$410 + 602 = 1012 \, \text{kJ mol}^{-1}$$

Hence, the hypothetical enthalpy of formation without resonance is

$$1190 - 1012 = 178 \, \text{kJ mol}^{-1}$$

Because resonance stabilizes the molecule, the actual enthalpy is lower than this expected value. Thus,

$$\text{Resonance energy} = 178 - 80 = 98 \, \text{kJ mol}^{-1}$$

Therefore, the required numerical value is 98.