Q16. 
Option(2). How?

​Q16. Given 
                
 Reaction  Energy Change (in kJ)
Li(s)  Li(g)
Li(g)  Li+(g)

1 2 F 2 g F g
F(g) + e¯ F¯(g)
Li+(g)  F¯(g)  LiF(s)
Li (s) +  1 2 F2(g)  Li F(s)
161
520


77

(Electron gain enthalpy)

–1047


–617
 
 
Based on data provided, the value of electron gain enthalpy of fluorine would be :

  (1) – 300 kJ mol–1                  (2) – 328 kJ mol–1                   (3) – 350 kJ mol–1                 (4) – 228 kJ mol–1
 

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