The bond dissociation enthalpy of $\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}$ calculated from the given…
$\begin{aligned}
& \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^*=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
$\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^*=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$
[Given : $\mathrm{M}^{+} \mathrm{X}^{-}$is a pure ionic compound and X forms a diatomic molecule $\mathrm{X}_2$ in gaseous state]
Solution
$\begin{aligned} & \begin{array}{l}\therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX})= \\ \quad \Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\text { I.E. }(\mathrm{M})+\frac{1}{2}[\text { B.E. }(\mathrm{X}-\mathrm{X})] \\ \\ +\mathrm{EG}(\mathrm{X})+\text { L.E. }(\mathrm{MX})\end{array} \\ & -400=(100)+(500)+\frac{1}{2}(\text { B.E. })+(-300)+(-800)\end{aligned} \quad \begin{aligned} & \therefore \text { B.E. }=200 \mathrm{~kJ} \mathrm{~mole}^{-1}\end{aligned}$
Asked in: JEE Main 2025 (23 Jan Shift 2)