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 $F$?

(N/A) Generally,when we move from left to right in the periodic table,the value of ionization enthalpy increases.
$(i)$ $\Delta_{i}H(Be) > \Delta_{i}H(B)$: $Be$ has a higher $\Delta_{i}H$ than $B$ because $Be$ has the electronic configuration $1s^2 2s^2$,while $B$ has the electronic configuration $1s^2 2s^2 2p^1$. The higher $\Delta_{i}H$ of $Be$ compared to $B$ is due to:
$\rightarrow$ The electronic configuration of $Be$ is more stable (completely filled $2s$ orbitals) than that of $B$.
$\rightarrow$ In $Be$,the electron to be removed is an $s$-electron,while in $B$,it is a $p$-electron.
$\rightarrow$ The penetration of a $2s$-electron to the nucleus is greater than that of a $2p$-electron. Consequently,the $2p$-electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$-electron of $Be$.
As a result,the $2s$-electron is more strongly attracted to the nucleus. Therefore,it is more difficult to remove a $2s$-electron from $Be$ than a $2p$-electron from $B$.
$(ii)$ $\Delta_{i}H(O) < \Delta_{i}H(N)$ and $\Delta_{i}H(O) < \Delta_{i}H(F)$: Oxygen has four electrons in $2p$-orbitals. Two of these electrons must occupy the same $2p$-orbital,resulting in increased electron-electron repulsion.
$\rightarrow$ Nitrogen $(N)$ has a stable half-filled configuration $(2p^3)$,which is highly stable.
$\rightarrow$ Fluorine $(F)$ has a higher nuclear charge than Oxygen,which holds its electrons more tightly.
Therefore,$O$ has a lower ionization enthalpy than both $N$ and $F$.