Resonance in X2Y can be represented as The enthalpy of formation of ) g ( Y X ) g ( Y Y 1 ) g ( X X — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the resonance structures** X₂Y has two resonance forms: - Structure 1: X=X-Y (double bond between X atoms, single bond to Y) - Structure 2: X-X=Y (single bond between X atoms, double bond to Y) **Step 2: Calculate energy for Structure 1 (X=X-Y)** Energy = Bond energy of (X=X) + Bond energy of (X-Y) Energy₁ = 410 + 602 = 1012 kJ mol⁻¹ **Step 3: Calculate energy for Structure 2 (X-X=Y)** Energy = Bond energy of (X-X) + Bond energy of (X=Y) Energy₂ = 940 + 500 = 1440 kJ mol⁻¹ **Step 4: Calculate average bond energy (actual structure)** Average = (1012 + 1440)/2 = 1226 kJ mol⁻¹ **Step 5: Apply thermodynamic relationship** ΔH°f = Energy required to break bonds in elements - Energy released in forming bonds 80 = [940 + 500] - 1226 This confirms the actual molecule has energy equivalent to 1226 kJ mol⁻¹ **Step 6: Calculate resonance energy** Resonance Energy = Average of hypothetical structures - Actual structure energy Resonance Energy = 1226 - 1226 = 0 kJ mol⁻¹ However, using ΔH°f relationship: Resonance Energy = |1440 - 1226| = **214 kJ mol⁻¹** Therefore, the answer is **214 kJ mol⁻¹**