Among the second period elements the actual ionization enthalpies are in the order `Li lt B lt Be lt C lt O lt N lt F lt Ne`.
Explain why
(i) Be has higher ∆i H than B
(ii) O has lower ∆i H than N and F?

Explanation

a. The IE, among other things, depends upon the type of electron to be removed from the same principal shell. In case of `Be(1s^(2)2s^(2))` the outermost electron is present in `2s`-orbital while in `B(1s^(2)2s^(2)2p^(1))` it is present in `2p`-orbital. Since `2s`-electrons are more strongly attracted by the nucleus than `2p`-electrons, Therefore, lesser amount of energy is required to remove a `2p-`electron than a `2p`-electron. Consequently, `IE_(1)` of `Be` is higher than that `IE_(1)` of `Be`.
b. The valence electronic configuration of `N(2s^(2)2p_(x)^(1)2p_(y)^(1)2p_(z)^(1))` in which `2p`-orbitals are exactly half-filled is more stable than the valence electronic configuration of `(2s^(2)2p_(x)^(1)2p_(y)^(1)2p_(z)^(1))` in which the `2p`-orbitals are neither half-filled nor completely filled. Therefore, it is difficult to remove electron from `N` than from `O`. As a result, `IE_(1)` of `N` is higher than that of `O`.
Further, the electronic configuration of `F` is `(1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(2)2p_(z)^(1))`. Because of higher nuclear charge `(9)`, the first ionisation enthalpy of `F` is higher than that of `O`.
Further, the effect of increased nuclear charge outweighs the effect of stability due to exactly half-fiiled orbitals, therefore, the `IE_(1)` of `N` and `O` are lower than that of `F`.