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Given Reaction Energy Change … Based on data provided, the value of electron gain enthalpy of fluorine would be :

JEE Main 2013 — Chemistry Physical Chemistry

2013mcqmedium

Given Reaction Energy Change  (in kJ) Li(s)→Li(g)161Li(g)→Li+(g)52012 F2( g)→F(g)77 F( g)+e−→F−(g) (Electron gain  enthalpy) Li+(g)+F−(g)→LiF(s)−1047Li(s)+12 F2( g)→LiF(s)−617\begin{array}{ll} & \text { (in kJ) } \\ \mathrm{Li}(\mathrm{s}) \rightarrow \mathrm{Li}(\mathrm{g}) & 161 \\ \mathrm{Li}(\mathrm{g}) \rightarrow \mathrm{Li}^{+}(\mathrm{g}) & 520 \\ \frac{1}{2} \mathrm{~F}_2(\mathrm{~g}) \rightarrow \mathrm{F}(\mathrm{g}) & 77 \\ \mathrm{~F}(\mathrm{~g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g}) & \begin{array}{l}\text { (Electron gain } \\ \text { enthalpy) }\end{array} \\ \mathrm{Li}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{LiF}(\mathrm{s}) & -1047 \\ \mathrm{Li}(\mathrm{s})+\frac{1}{2} \mathrm{~F}_2(\mathrm{~g}) \rightarrow \mathrm{LiF}(\mathrm{s}) & -617\end{array}Li(s)→Li(g)Li(g)→Li+(g)21​ F2​( g)→F(g) F( g)+e−→F−(g)Li+(g)+F−(g)→LiF(s)Li(s)+21​ F2​( g)→LiF(s)​ (in kJ) 16152077 (Electron gain  enthalpy) ​−1047−617​ Based on data provided, the value of electron gain enthalpy of fluorine would be :

Official previous-year question

Held on 22 Apr 2013 · Verified 6 Jul 2026.

Options

  1. A

    −300 kJ mol−1-300 \mathrm{~kJ} \mathrm{~mol}^{-1}−300 kJ mol−1

  2. B

    −350 kJ mol−1-350 \mathrm{~kJ} \mathrm{~mol}^{-1}−350 kJ mol−1

  3. C

    −328 kJ mol−1-328 \mathrm{~kJ} \mathrm{~mol}^{-1}−328 kJ mol−1

  4. D

    −228 kJ mol−1-228 \mathrm{~kJ} \mathrm{~mol}^{-1}−228 kJ mol−1

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