Resonance in $\mathrm{X}_2\mathrm{Y}$ can be represented as (see structure) The enthalpy of formation of $\mathrm{X}_2\mathrm{Y}$ ($\mathrm{X}\equiv\mathrm{X}(\mathrm{g}) + \frac{1}{2}\mathrm{Y}=\mathrm{Y}(\mathrm{g}) \rightarrow \mathrm{X}_2\mathrm{Y}(\mathrm{g})$) is 80 kJ mol⁻¹. The magnitude of resonance energy of $\mathrm{X}_2\mathrm{Y}$ is _____ kJ mol⁻¹ (nearest integer value) Given: Bond energies of $\mathrm{X}\equiv\mathrm{X}$, $\mathrm{X}=\mathrm{X}$, $\mathrm{Y}=\mathrm{Y}$ and $\mathrm{X}=\mathrm{Y}$ are 940, 410, 500 and 602 kJ mol⁻¹ respectively. valence X: 3, Y: 2

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Chemical Thermodynamics

Resonance in

X_{2} Y

can be represented as

(see structure)

The enthalpy of formation of

X_{2} Y

(

X \equiv X (g) + \frac{1}{2} Y = Y (g) \to X_{2} Y (g)

) is 80 kJ mol⁻¹.

The magnitude of resonance energy of

X_{2} Y

is _____ kJ mol⁻¹ (nearest integer value)

Given: Bond energies of

X \equiv X

X = X

Y = Y

and

X = Y

are 940, 410, 500 and 602 kJ mol⁻¹ respectively.
valence X: 3, Y: 2

Answer: 98

Solution

Resonance energy is the extra stability by which the measured enthalpy of formation lies below the value estimated from a single Lewis structure. Taking the drawn structure

X = X = Y

, which contains one

X = X

bond and one

X = Y

bond, the estimated enthalpy of formation is

Δ H_{f}^{(struct)} = [E_{X \equiv X} + \frac{1}{2} E_{Y = Y}] - [E_{X = X} + E_{X = Y}] = [940 + 250] - [410 + 602] = 178 kJ mol^{- 1}

. Since the experimental enthalpy of formation is

80 kJ mol^{- 1}

, the resonance energy is

80 - 178 = - 98 kJ mol^{- 1}

. The magnitude of the resonance energy is

98 kJ mol^{- 1}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Chemical Thermodynamics

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