For $NaCl (s)$ : $Δ_{f} H^{\circ} = - 411$ , sublimation of $Na (s) = + 108$ , ionisation enthalpy of $Na (g) = + 496$ , $Cl - Cl$ bond enthalpy $= + 242$ , electron gain enthalpy of $Cl (g) = - 349$ (all in $kJ mo l^{- 1}$ ). The lattice enthalpy for $Na X^{+} (g) + Cl X^{-} (g) NaCl (s)$ is

Answer: C

Born-Haber cycle:

Δ_{f} H = Δ_{s u b} H + I E + \frac{1}{2} BE + E A + Δ_{l a t} H

. The energy to make the gaseous ions is

108 + 496 + 121 - 349 = 376 kJ

. So

Δ_{l a t} H = - 411 - 376 = - 787 kJ mo l^{- 1}

(the dissociation of only half a mole of

Cl X_{2}

is needed).