Given
Reaction Energy Change (in kJ) \(\mathrm{Li(s)} \to \mathrm{Li(g)}\) 161 \(\mathrm{Li(g)} \to \mathrm{Li^+(g)}\) 520 \(\frac{1}{2}\mathrm{F_2(g)} \to \mathrm{F(g)}\) 77 \(\mathrm{F(g)} + \mathrm{e^-} \to \mathrm{F^-(g)}\) (Electron gain enthalpy) \(\mathrm{Li^+(g)} + \mathrm{F^-(g)} \to \mathrm{LiF(s)}\) \(-1047\) \(\mathrm{Li(s)} + \frac{1}{2}\mathrm{F_2(g)} \to \mathrm{LiF(s)}\) \(-617\)
Based on data provided, the value of electron gain enthalpy of fluorine would be :
| Reaction | Energy Change (in kJ) |
| \(\mathrm{Li(s)} \to \mathrm{Li(g)}\) | 161 |
| \(\mathrm{Li(g)} \to \mathrm{Li^+(g)}\) | 520 |
| \(\frac{1}{2}\mathrm{F_2(g)} \to \mathrm{F(g)}\) | 77 |
| \(\mathrm{F(g)} + \mathrm{e^-} \to \mathrm{F^-(g)}\) | (Electron gain enthalpy) |
| \(\mathrm{Li^+(g)} + \mathrm{F^-(g)} \to \mathrm{LiF(s)}\) | \(-1047\) |
| \(\mathrm{Li(s)} + \frac{1}{2}\mathrm{F_2(g)} \to \mathrm{LiF(s)}\) | \(-617\) |
Show Hint
- \(-300\ \mathrm{kJ\ mol^{-1}}\)
- \(-350\ \mathrm{kJ\ mol^{-1}}\)
- \(-328\ \mathrm{kJ\ mol^{-1}}\)
- \(-228\ \mathrm{kJ\ mol^{-1}}\)
The Correct Option is C
Solution and Explanation
\(-617 = 161 + 520 + 77 + \Delta H_{eg} + (-1047)\).
\(161 + 520 + 77 - 1047 = -289\).
\(\Delta H_{eg} = -617 + 289 = -328\ \mathrm{kJ\ mol^{-1}}\).
Answer: (3) \(-328\ \mathrm{kJ\ mol^{-1}}\)
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