Prove that the formation of `NaCI_(2)` is highly endothermic and energetically unfavorable from the following data using Born-Haber Cycle
`{:("Heat of sublimation of"Na,=,109kJ//mol),(1^(st)IE of Na,=,494 kJ//mol),(1^(st)IE of Na^(+),=,4563 kJ//mol),("Bond Energy of" CI_(2),=,242 kJ//mol),("Electron gain enthalpy of"CI,=,-347 kJ//mol),("Lattice energy of" NaCI_(2),=,2560 kJ//mol):}`

Answer

`[2154 kJ//mol]`

Explanation

`DeltaH_(f) = (DeltaH)_("sub")` of `Na +(IE_(1) +IE_(2))` of `Na + BDE(CI_(2)) +2EA_(1) (CI)+U`
After putting the value `(DeltaH)` value is obtain with positive significance. Therefore formation of `NaCI_(2)` is highly endothermcic & energetically unfavorable