For each of the following pairs predict which one has lower first ionization enthalpy ?
(i) N or O (ii) Na or `Na^(+)` (iii) `Ba^(+) "or " Mg^(2+) (iv) " or " I^(-)`

Answer

(i) O, (ii) Na , (iii) `Be^(+)` (iv) I

Explanation

(i) O has lower `Delta_(i) H_(1)` than N because in case of O , loss of an electron gives `O^(+)` which has stable exactly half- filled electronic configuration while in case of N, an electron has to the to be lost from a stable half- filled electronic configuration
(ii) Na has lower `Delta_(i) H_(1)` than `Na^(+)` because of the following two reasons : (a) In case of `Na^(+)` an electron has to be lost from a stable inert gas configuration but in case of Na , loss of an electron gives stable inert gas configuration (b) Na is neutral but `Na^(+)` is + vely charged
(iii) `Be^(+)` has lower `Delta_(i)H_1)` than `Mg^(2+)` because in case of `Be^(+)` the loss of one electron gives a stable inert gas configuration but in case of `Mg^(2+)` the electron has to be lost from the stable inert gas configuration .
(iv) I has lower `Delta_(i) H_(1)` than `I^(-)` because in case of `I^(-)` an electron has to be lost from a stable inert gas configuration