Question:
Given the following data:
l c
Reaction & Energy change (kJ)
Li(s) → Li(g) & 161
Li(g) → Li⁺(g) & 520
(1)/(2)F₂(g) → F(g) & 77
F(g) + e⁻ → F⁻(g) & ?
Li⁺(g) + F⁻(g) → LiF(s) & -1047
Li(s) + (1)/(2)F₂(g) → LiF(s) & -617
Based on the data provided, the value of electron gain enthalpy of fluorine would be:
Given the following data:
l c
Reaction & Energy change (kJ)
Li(s) → Li(g) & 161
Li(g) → Li⁺(g) & 520
(1)/(2)F₂(g) → F(g) & 77
F(g) + e⁻ → F⁻(g) & ?
Li⁺(g) + F⁻(g) → LiF(s) & -1047
Li(s) + (1)/(2)F₂(g) → LiF(s) & -617 Based on the data provided, the value of electron gain enthalpy of fluorine would be:
Li(s) → Li(g) & 161
Li(g) → Li⁺(g) & 520
(1)/(2)F₂(g) → F(g) & 77
F(g) + e⁻ → F⁻(g) & ?
Li⁺(g) + F⁻(g) → LiF(s) & -1047
Li(s) + (1)/(2)F₂(g) → LiF(s) & -617 Based on the data provided, the value of electron gain enthalpy of fluorine would be:
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Electron affinity can be calculated using Born–Haber cycle.
Updated On: Mar 20, 2026
- -300kJ mol⁻1
- -350kJ mol⁻1
- -328kJ mol⁻1
- -228kJ mol⁻1
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The Correct Option is C
Solution and Explanation
Step 1: Apply Hess’s law for formation of LiF(s): 161+520+77+EA -1047 = -617
Step 2: Simplifying: 758 + EA -1047 = -617 ⟹ EA = -328kJ mol⁻1
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