Chemistry · 2013
JEE Main · 22 April 2013, Shift 1 · Q33
Based on data provided, the value of electron gain enthalpy of fluorine would be:
| Given | |
| Reaction | Energy Change (in kJ) |
| Li (s) → Li (g) | $\displaystyle 161$ |
| $\displaystyle \mathrm{Li}(\mathrm{g}) \rightarrow \mathrm{Li}^{+}(\mathrm{g})$ | $\displaystyle 520$ |
| $\displaystyle \frac{1}{2} \mathrm{~F}_2(\mathrm{~g}) \rightarrow \mathrm{F}(\mathrm{g})$ | $\displaystyle 77$ |
| $\displaystyle \mathrm{F}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g})$ | (Electron gain enthalpy) |
| $\displaystyle \mathrm{Li}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{Li} \mathrm{F}(\mathrm{s})$ | -$\displaystyle 1047$ |
| $\displaystyle \mathrm{Li}(\mathrm{s})+1 / 2 \mathrm{~F}_2(\mathrm{~g}) \rightarrow \mathrm{LiF}(\mathrm{s})$ | -$\displaystyle 617$ |
Official answer
(3)
$\displaystyle -328 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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JEE Main 2013 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.