Applying Hess's Law ΔfH∘=Δsub H+21Δdiss H+ I.E. + E.A +Δlattice H−617=161+520+77+ E.A. +(−1047) E.A. =−617+289=−328 kJ mol−1∴ electron affinity of fluorine =−328 kJ mol−1
JEE Main 2013 — Chemistry Physical Chemistry
Given Reaction Energy Change 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 :
Held on 22 Apr 2013 · Verified 6 Jul 2026.
−300 kJ mol−1
−350 kJ mol−1
−328 kJ mol−1
−228 kJ mol−1
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