Question

Resonance in X$_2$Y can be represented as

The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is:

Hint:

To calculate resonance energy, subtract the enthalpy of formation from the bond dissociation enthalpy. Be sure to account for the stoichiometric coefficients when combining bond energies.

Answer 1

Correct Answer: 98

Step 1: Determine the expected enthalpy of formation using bond energies.
- Break \( X = X \) bond: \( +940 \, \text{kJ mol}^{-1} \).
- Break \( \frac{1}{2} Y = Y \) bond: \( +\frac{1}{2} \times 500 = 250 \, \text{kJ mol}^{-1} \).
- Total energy input = \( 940 + 250 = 1190 \, \text{kJ mol}^{-1} \).

Step 2: Energy released when forming bonds in \( X_2Y \).
- Assume one \( X - X \) bond (\( 410 \, \text{kJ mol}^{-1} \)) and one \( X - Y \) bond (\( 602 \, \text{kJ mol}^{-1} \)).
- Total energy released = \( 410 + 602 = 1012 \, \text{kJ mol}^{-1} \).

Step 3: Calculate expected enthalpy change.
- Expected \( \Delta H = 1190 - 1012 = 178 \, \text{kJ mol}^{-1} \).

Step 4: Calculate resonance energy.
- Given enthalpy of formation = \( 80 \, \text{kJ mol}^{-1} \).
- Resonance energy = Expected \( \Delta H \) - Actual \( \Delta H \).
- Resonance energy = \( 178 - 80 = 98 \, \text{kJ mol}^{-1} \).

Step 5: Verify. - The resonance structures suggest partial triple bond character, 
stabilizing the molecule, aligning with the calculated value.
The magnitude of the resonance energy, to the nearest integer, is \( 98 \, \text{kJ mol}^{-1} \).