(i) The ionization enthalpy, among other things, depends upon the type of electron to be removed from the same principal shell. In case of \(\mathrm{Be}\left(1 s^2 2 s^2\right)\) the outermost electron is present in \(2 s\)-orbital while in B \(\left(1 s^2 2 s^2 2 p^1\right)\) it is present in \(2 p\)-orbital. Since \(2 s\) electrons are more strongly attracted by the nucleus
than \(2 p\)-electrons, therefore, lesser amount of energy is required to knock out a \(2 p\)-electron than a \(2 s\) electron. Consequently, \(\Delta_{\mathrm{i}} \mathrm{H}\) of \(\mathrm{Be}\) is higher than that \(\Delta_{\mathrm{i}} \mathrm{H}\) of \(\mathrm{B}\).
(ii) The electronic configuration of \(\mathrm{N}\left(1 s^2 2 s^2 2 p_x{ }^1 2 p_y{ }^1\right.\) \(2 p_z^1\) ) in which \(2 p\)-orbitals are exactly half-filled is more stable than the electronic configuration of \(\mathrm{O}\left(1 s^2\right.\) \(2 s^2 2 p_x^2, 2 p_y^1, 2 p_z^1\) ) in which the \(2 p\)-orbitals are neither half-filled nor completely filled. Therefore, it is difficult to remove an electron from \(\mathrm{N}\) than from
\(\mathrm{O}\). As a result, \(\Delta_{\mathrm{i}} \mathrm{H}\) of \(\mathrm{N}\) is higher than that of \(\mathrm{O}\). Further, the electronic configuration of \(\mathrm{F}\) is \(1 s^2 2 s^2\) \(2 p_x^2 2 p_y^2 2 p_z^1\). Because of higher nuclear charge \((+9)\), the first ionization enthalpy of \(\mathrm{F}\) is higher than that of \(\mathrm{O}\).