Calculate the molar enthalpy of formation of solid lithium fluoride

Answer  : The molar enthalpy of formation of LiF will be -607 KJ.

Explanation : In the given reaction of 



we have to use hess's law of summation in this reaction, which is attached in the attachment;

we have got,
  ΔU = -1037 KJ/mol
= 159.37 KJ/mol
IE = 520 KJ/mol
  = 78.99 KJ/mol
EA = -328.0 KJ/mol

Hence ΔH(formation) = ΔU+++EA+IE

ThereforeΔH(formation) = -1037 + 159.37 + 78.99 + -328 + 520
= -607 KJ/mol

So, the enthalpy of formation of LiF will be as -607 KJ/mol

Outermost (valence) electrons