JEE MAIN - Chemistry (2025 - 8th April Evening Shift - No. 25)
Resonance in $\mathrm{X}_2 \mathrm{Y}$ can be represented as
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The enthalpy of formation of $X_2Y$ $ \left(X = X(g) + \frac{1}{2} Y = Y(g) \rightarrow X_2Y(g) \right) $ is 80 kJ mol$^{-1}$. The magnitude of resonance energy of $X_2Y$ 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
Explanation
$$\Delta \mathrm{H}_{\mathrm{R} . \mathrm{E}}=\Delta \mathrm{H}_{\mathrm{f}(\mathrm{exp})}-\Delta \mathrm{H}_{\mathrm{f}(\text { Theo })}$$
$\Delta \mathrm{H}_{\mathrm{f}(\mathrm{exp})}$ for $\mathrm{X}_2 \mathrm{Y}_{(\mathrm{g})}=80 \mathrm{~kJ} / \mathrm{mole}$
for $\Delta \mathrm{H}_{\mathrm{f} \text { (Theo) }}$
$$\mathrm{X}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Y}_{2(\mathrm{~g})} \rightarrow \mathrm{X}_2 \mathrm{Y}_{(\mathrm{g})} \Delta \mathrm{H}_{\mathrm{f}}=?$$
$$ \begin{aligned} & \Delta \mathrm{H}_{\mathrm{f}(\text { Theo })}=\left(\mathrm{BE}_{\mathrm{X}=\mathrm{X}}+\frac{1}{2} \mathrm{BE}_{\mathrm{Y}=\mathrm{Y}}\right)-\left(\mathrm{BE}_{\mathrm{X}=\mathrm{X}}+\mathrm{BE}_{\mathrm{X}=\mathrm{Y}}\right) \\ & =\left(940+\frac{1}{2} \times 500\right)-(410+602) \\ & =178 \mathrm{~kJ} / \mathrm{mole} \\ & \Delta \mathrm{H}_{\text {R.E }}=80-178 \\ & \quad=-98 \mathrm{~kJ} / \mathrm{mol} \\ & |\Delta \mathrm{H}_{\text {R.E }}|=98 \end{aligned}$$
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