For the reaction `K(g)+F(g) +K^(o+)+F^(Theta)` (separated ions `DeltaH = 19 kcal mol^(-1)`), if the ionisation potential of `K` and the electron affinity of `F^(Theta)` have a geometric means of `3.88eV` and `IP gt EA`, calculate the values fo ionisation potential and electron affinity.

Explanation

`K_((g)) rarrK_((g))^(o+)+e^(-),IE = x kcal mol^(-1)` ....(i)
`F_((g))+e^(-) rarrF_((g))^(ɵ),Delta_("eg")H^(ɵ) = -y kcal mol^(-1)` ...(ii)
Adding Eqs. (i) and (ii), we get
`K_((g))+F_((g)) rarrK_((g))^(o+)+F_((g))^(ɵ)`
`:. x-y = 19 kcal mol^(-1) = 0.82 eV` ...(iii)
`sqrt(xy) = 3.88 eV rArr xy = (3.88)^(2) = 15.0544`
`:. x = (15.0544)/(y)`
Substituting the value of `x` in Eq. (iii), we get
`(15.0544)/(y)-y = 0`
Solving, we get
`y = 3.48 eV`
`x+y = 7.48 eV`, `x-y = 0.82 eV`
`:. IE = x = 4.313 eV`
`Delta_("eg")H^(ɵ) = -y = -3.48 eV`