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Q.

Given 

ReactionEnergy change (in Kj)
Li(s)Li(g)161
Li(g)Li+(g)520
12F2(g)F(g)77
F(g)+eF(g)(Electron gain enthalpy)
Li+(g)+F(g)LiF(s)-1047
Li(s)+12F2(g)LiF(s) -617
Based on data provided, the value of electron gain enthalpy of fluorine would be xkjmol1. X value 

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answer is 328.

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Detailed Solution

From Born-Haber cycle, 
Q=S+I+D+EA+U617=161+520+77+EA1047
,Here,S=  Sublimation    energy,I=ionisation     energy
D=dissociation energy, EA =electron gain enthalpy
And U = lattice energy]
 EA=289617=328kJ  mol1
 

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