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The bond dissociation enthalpy of \(\mathrm{X}_{2} \ \Delta \mathrm{H}_{\mathrm{bond}}^{\circ}\) calculated from the given data is _____ \(\mathrm{kJ} \ \mathrm{mol}{ }^{-1}\).

(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]

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Best answer

Answer is : 200

The bond dissociation 

\(\therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX})=\Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\) I.E. \((\mathrm{M})+\frac{1}{2}[\) B.E. \((\mathrm{X}-\mathrm{X})]\) \(+\ \mathrm{EG}(\mathrm{X})+\) L.E.(MX)

\( -400=(100)+(500)+\frac{1}{2}( B.E. )+(-300)+(-800)\) 

\( \therefore\ B.E. =200 \mathrm{~kJ}\ mole ^{-1}\)

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