The Born-Haber cycle relates the standard enthalpy of formation ($\Delta H_f^\circ$) of an ionic compound to various thermodynamic quantities, including lattice enthalpy ($\Delta H_{\text{lattice}}^\circ$). The cycle allows us to calculate the lattice enthalpy using the following relationship:
$ \Delta H_f^\circ(\text{LiF}) = \Delta H_{\text{sub}}^\circ(\text{Li}) + \Delta H_{IE}^\circ(\text{Li}) + \frac{1}{2} \Delta H_{\text{diss}}^\circ(\text{F}_2) + \Delta H_{\text{EGE}}^\circ(\text{F}) + \Delta H_{\text{lattice}}^\circ(\text{LiF}) $
We need to find the lattice enthalpy ($\Delta H_{\text{lattice}}^\circ(\text{LiF})$). Rearranging the Born-Haber equation:
$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = \Delta H_f^\circ(\text{LiF}) - \Delta H_{\text{sub}}^\circ(\text{Li}) - \Delta H_{IE}^\circ(\text{Li}) - \frac{1}{2} \Delta H_{\text{diss}}^\circ(\text{F}_2) - \Delta H_{\text{EGE}}^\circ(\text{F}) $
Substitute the provided enthalpy values into the rearranged equation:
Calculation:
$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = (-594) - (155) - (520) - (75) - (-313) $
$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = -594 - 155 - 520 - 75 + 313 $
$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = -1344 + 313 $
$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = -1031 \text{ kJ mol}^{-1} $
The question asks for the magnitude of the lattice enthalpy.
Magnitude = $ |-1031 \text{ kJ mol}^{-1}| = 1031 \text{ kJ mol}^{-1} $
The magnitude of the lattice enthalpy of LiF is 1031 kJ mol⁻¹.
Given below are two statements :
Statement I : $\text{C} < \text{O} < \text{N} < \text{F}$ is the correct order in terms of first ionization enthalpy values.
Statement II : $\text{S} > \text{Se} > \text{Te} > \text{Po} > \text{O}$ is the correct order in terms of the magnitude of electron gain enthalpy values.
In the light of the above statements, choose the correct answer from the options given below :