Q30 / 125 · BITSAT 2016 · Chemistry
Given
$\mathrm{F}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g}) \quad$ (Electron gain enthalpy)
$\mathrm{Li}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{LiF}(\mathrm{s}) \quad-1047$
$\mathrm{Li}(\mathrm{s})+\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{LiF}(\mathrm{s}) \quad-617$
Based on data provided, the value of electron gain enthalpy of fluorine would be:
$\mathrm{F}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g}) \quad$ (Electron gain enthalpy) $\mathrm{Li}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{LiF}(\mathrm{s}) \quad-1047$ $\mathrm{Li}(\mathrm{s})+\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{LiF}(\mathrm{s}) \quad-617$ Based on data provided, the value of electron gain enthalpy of fluorine would be:
- A$-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$-228 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Applying Hess's Law
$$
\begin{array}{c}
\Delta_{\mathrm{f}} \mathrm{H}^{\circ}=\Delta_{\mathrm{sub}} \mathrm{H}+\frac{1}{2} \Delta_{\mathrm{diss}} \mathrm{H}+\mathrm{L} \text { .E. }+\mathrm{E} . \mathrm{A}+\Delta_{\text {lattice }} \mathrm{H} \\
-617=161+520+77+\mathrm{E} \mathrm{A} .+(-1047) \\
\mathrm{E} . \mathrm{A} .=-617+289=-328 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{array}
$$
$\therefore$ electron affinity of fluorine $=-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Hence, the correct answer is option: 'C'.