Answer the following question (Based on `EA, Delta_(eg)H^(ɵ)` and `IE`).
(a) `IE_(1)` of `Li` is `5.4 eV "atom"^(-1)` and the `EA` of `Cl` is `3.6 eV "atom"^(-1)`. Calculate `Delta_( r)H^(ɵ)` in `"kcal mol"^(-1)` and `kJ mol^(-1)` for the reaction
`Li_((g))+Cl_((g)) rarr Li_((g))^(o+)+Cl_((g))^(ɵ)`
formed at such a low pressure that resulting ions do not combine with each other.
(b) The `IE` of atoms `X` and `Y` are `400` and `300 kcal mol^(-1)` respectively. `EA's` of these atoms are `80.0` and `85.0 K cal mol^(-1)`. Explain as which of the atoms has higher `EN`.
( c) Explain why `EA` of `S` is `-200 kJ mol^(-1)` but the second `EA` is `+649 kJ mol^(-1)`?
(d) Which of the following pairs of elements would have more negative electron gain enthalpy `(Delta_("eg")H^(ɵ))` ?
(i) `F` or `Cl` , (ii) `O` or `F`
(e) What would be the second electron gain enthalpy `(Delta_(eg)H_(2)^(ɵ))` of oxygen as positive, more negative or less negative than first `(Delta_(eg)H_(1)^(ɵ))`? Explain.
(f) Which has less negative `Delta_(eg)H^(ɵ)` oxygen or sulphur?

Explanation

(a) `Li_((g))rarrLi_((g))^(o+)+e^(-)`, `IE_(1) = 5.4 eV "atom"^(-1)` …(i)
`Cl_((g))+e^(-)Cl_((g))^(ɵ)`, `EA_(1) = -3.6 eV = -3.6 eV "atom"^(-1)` …(ii)
Note: `EA_(1)` is negative.
Adding Eqs. (i) and (ii), we get
`Li_((g))+Cl_((g)) rarr Li_((g))^(o+)+Cl_((g))^(ɵ)`
`:. Delta_( r)H^(ɵ) = IE_(1) + EA_(1) = 5.4 + (-3.6) = 1.8 eV "atom"^(-1)`
We know `1 eV "atom"^(-1) = 23.06 kcal mol^(-1)` and `1 eV "atom"^(-1) = 96.49 kJ mol^(-1)`
`:. Delta_(r )H^(ɵ)` (in `kJ mol^(-1)`) `= 1.8 xx 23.06`
`= 41.508 kJ mol^(-1)`
`:. Delta_(r )H^(ɵ)` (in `kcal mol^(-1)`) `= 1.8 xx 96.49`
`= 173.682 kcal mol^(-1)`
(b) `EN` of `X = (IE+EA)/(125)` (when `IE` and `EA` values are in `kcal mol^(-1)`)
`= (400+80)/(125) = 3.84`
`EN` of `Y = (300+85)/(125) = 3.08`
`:. EN` of `X gt EN` of `Y`
(e) `S_((g))+e^(-)rarrS_((g))^(ɵ), EA_(1) = -200 kJ mol^(-1)`
`S_((g))^(ɵ)+e^(-) rarrS_((g))^(2-), EA_(2) +649 kJ mol^(-1)`
The addition of the second electron to `S^(ɵ)(g)` experience repulsion from the already present electrons.
(d) i. `Cl`. When electron is added to `F`, the added electron goes to smaller `n = 2` (i.e., `2p` orbital) quantum level and suffers significant repulsion from the other electrons present in the shell. So, electron is being added with difficulty in `F` than in `Cl`. Hence less negative energy in released in `F` than in `Cl`. So, `Delta_(eg)H^(ɵ)` of `F` is less negative than `Cl`.
(ii) `F. Delta_(eg)H^(ɵ)` of `F` is more negative then `O` due to the small size of `F` than `O`. Moreover, `Delta_(eg)H^(ɵ)` increases (more negative) along the period `(rarr)`.
(e) `e^(-)+O_((g)) rarr O_((g))^(ɵ) EA_(1) = -200 kJ mol^(-1)`
`e^(-)+O_((g))^(ɵ) rarrO_((g))^(2-), EA_(2)=?`
`EA_(2)` is positive sue to move electrons-electron repulsion.
(f) Due to the compact nature of oxygen atom it has less negative `Delta_(eg)H^(ɵ)(-14 kJ mol^(-1))` than `S(-200 kJ mol^(-1))`. [Same explanation as in `Delta_(eg)H^(ɵ)` of `F` and `Cl`].