From `N` atoms of an element `A`, when half the atoms transfer one electron to the another atom. `405 kJ mol^(-1)` of energy was found to be consumed. An additional energy of `745 kJ mol^(-1)` was further required to convert all the `A^(ɵ)` ions to `A^(o+)`. Calculate the ionisation energy and the electron gain enthalpy of atom `A` in `eV (1 eV = 96.48 kJ)`.

Explanation

`N//2(ArarrA^(o+)+e^(-))`, `IE = +ve`
`N//2(A+e^(-) rarr A^(ɵ)), Delta_("eg")H^(ɵ)-ve`
Let `x` and `y` are the `IE` and `Delta_("eg")H^(ɵ)` respectively.
`:. (1)/(2)(x-y) = 405 kJ`
`(1)/(2)(x+y) = 745 kJ`
On solving `x = 1150` and `y = 340 kJ`
or `1150 kJ = 11.91 eV`
Similarly, `340 kJ = 3.52 eV`
`IE = 11.9 eV` and `Delta_("eg")H^(ɵ) = -3.5 eV`