JEE MAIN - Chemistry (2025 - 23rd January Evening Shift - No. 22)
The bond dissociation enthalpy of $\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}^{\circ}$ calculated from the given data is ___________ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)
$$\begin{aligned} & \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^{\circ}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$
$$\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
$$\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
$$\mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
[Given : $\mathrm{M}^{+} \mathrm{X}^{-}$is a pure ionic compound and X forms a diatomic molecule $\mathrm{X}_2$ in gaseous state]
Explanation
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$$\begin{aligned} & \begin{aligned} \therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX}) & =\Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\text { I.E. }(\mathrm{M})+\frac{1}{2}[\text { B.E. }(\mathrm{X}-\mathrm{X})] \\ & +\mathrm{EG}(\mathrm{X})+\text { L.E. }(\mathrm{MX}) \end{aligned} \\ & -400=(100)+(500)+\frac{1}{2}(\text { B.E. })+(-300)+(-800)\\ & \therefore \text { B.E. }=200 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned} ~ \begin{aligned} \end{aligned}$$
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