Resonance in X₂ Y can be represented as
Ẍ(⊖) = X = Ÿ(⊕) ↔ :X ≡ X(⊕) - Ÿ(⊖):
The enthalpy of formation of X₂Y (X = X(g) + 1/2 Y = Y(g) → X₂Y(g) ) is 80 kJ mol⁻¹. The magnitude of resonance energy of X₂Y is __ kJ mol⁻¹ (nearest integer value).
Given: Bond energies of X ≡ X, X = X, Y = Y and X = Y are 940, 410, 500, and 602 kJ mol⁻¹ respectively.
valence X: 3, Y: 2

Solution

Chemistry ma’am · 4 steps

Find ΔH_f from bond energies of one resonance structure; the real molecule's lower value is resonance energy.

1. Read the reactants. X has valence 3, so the X₂ molecule is X≡X (940). Y has valence 2, so Y₂ is Y=Y (500). The given equation prints "X = X", but the valencies tell you the reactant has a triple bond.
2. Pick a structure. Take the first resonance structure, X=X=Y. It has one X=X bond (410) and one X=Y bond (602), and both bond energies are given.

   Δ H_(calc) = (940 + 1/2× 500) - (410 + 602) = 1190 - 1012 = 178 kJ mol⁻¹

3. Compare with actual. The real X₂Y has Δ H_f = 80 kJ mol⁻¹. It is more stable than any single structure, so its value is lower. Resonance energy = 178 - 80.
4. So. The magnitude of the resonance energy is 98 kJ mol⁻¹.

The common slip: 432. Taking the reactant X₂ as X=X (410) instead of X≡X (940) gives Δ H_(calc) = -352 and a resonance energy of 432, not 98.

The idea to keep. Resonance energy is ΔH from bond energies of one structure minus the experimental ΔH. Bonds broken minus bonds formed, and work out the bond orders from the valencies.

Answer from the NTA final answer key. Solution by PrepWiser.