Among the second period elements,the actual ionization enthalpies are in the order $Li < B < Be < C < O < N < F < Ne$. Explain why:
$(i)$ $Be$ has higher $\Delta_{i}H$ than $B$.
$(ii)$ $O$ has lower $\Delta_{i}H$ than $N$ and $F$.

(A) $(i)$ During the process of ionization,the electron to be removed from a beryllium atom is a $2s$-electron,whereas the electron to be removed from a boron atom is a $2p$-electron.
$2s$-electrons are more strongly attracted to the nucleus than $2p$-electrons. Therefore,more energy is required to remove a $2s$-electron of beryllium than that required to remove a $2p$-electron of boron. Hence,beryllium has a higher $\Delta_{i}H$ than boron.
$(ii)$ In nitrogen $(1s^2 2s^2 2p^3)$,the three $2p$-electrons occupy three different atomic orbitals,representing a stable half-filled configuration. In oxygen $(1s^2 2s^2 2p^4)$,two of the four $2p$-electrons occupy the same $2p$-orbital,resulting in increased electron-electron repulsion. Consequently,the energy required to remove an electron from oxygen is less than that required for nitrogen.
Fluorine has a higher effective nuclear charge than oxygen due to an additional proton in the same shell. This results in a stronger attraction for valence electrons,making it harder to remove an electron from fluorine compared to oxygen. Thus,oxygen has a lower $\Delta_{i}H$ than both nitrogen and fluorine.