Chemistry · 2025
JEE Main · 23 January 2025, Shift 2 · Q72
The bond dissociation enthalpy of X_2 Δ H_bond^° calculated from the given data is ____ kJ mol^-1. (Nearest integer) M^+ X^-( s ) → M^+( g )+ X^-( g…
The bond dissociation enthalpy of $\displaystyle \mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}^{\circ}$ calculated from the given data is $\displaystyle \_\_\_\_$
$\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)
$$\begin{aligned}
& \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^{\circ}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
[Given : $\displaystyle \mathrm{M}^{+} \mathrm{X}^{-}$is a pure ionic compound and X forms a diatomic molecule $\displaystyle \mathrm{X}_2$ in gaseous state]
Official answer
From NTA’s final answer key for this paper.
200
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JEE Main 2025 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.