Ionisation of energy `F^(ɵ)` is `320 kJ mol^(-1)`. The electronic gain enthalpy of fluorine would be

Ionisation of energy of F^(-) is 320 kJ mol^(-1) . The electron gain enthalpy of fluorine would be

Ionisation energy of F^Ɵ is 320 kJ mol^(-1) . The electron gain enthalpy of fluorine would be

Ionisation energy of F is 320 kJ mol^(-1) . The electron affinity of fluorine would be:

The formation of O^(2-)(g) startng from O(g) is endothermic by 603 kJ mol^(-1) . If electron gain enthalpy of O(g) is -141 kJ mol^(-1) , the second electron gain enthalpy of oxygen would be :

Given Based on data provided, the value of electron gain enthalpy of fluorine would be

Use the following data to calculate Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ) for sodium metal = 108.4 kJ mol^(-1) , ionisation enthalpy of sodium = 496 kJ mol^(-1) , electron gain enthalpy of bromine = -325 kJ mol^(-1) , bond dissociation enthalpy of bromine = 192 kJ mol^(-1), Delta_(f) H^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)