JEE Main 2025 — 23 January, Evening Shift — Question 46

The bond dissociation enthalpy of $\mathrm{X}_{2} \Delta \mathrm{H}_{\mathrm{bond}}^{\circ}$

calculated from the given data is $\qquad$ $\mathrm{kJ} \mathrm{mol}{ }^{-1}$. (Nearest integer)

$\mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^{\circ}=800 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{M}(\mathrm{s}) \rightarrow \mathrm{M}(\mathrm{g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})^{-}+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{X}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}^{\circ}{ }_{\mathrm{eg}}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{M}(\mathrm{s})+\frac{1}{2} \mathrm{X}_{2}(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$

[Given : $\mathrm{M}^{+} \mathrm{X}^{-}$is a pure ionic compound and X forms a diatomic molecule $\mathrm{X}_{2}$ is gaseous state]