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 ${\Delta }_{I.E}H$ than $B$ (ii) $O$ has lower ${\Delta }_{I.E}H$ than $N$ and $F$?

Half filled orbitals and fully filled orbitals are more stable.

$\left(i\right)$ Atomic no. of $Be=4\to 1s^{2}2s^2$

Atomic no. of $B=5\to 1s^{2}2s^{2}2p^1$

In case of $Be$ it has fully filled $s-$orbital thus it is more stable and more ionization enthalpy is needed to remove an electron from it.

In case of $B$, the outermost electron is in $p-$orbital , after removal of which it become stable with stable configuration $1s^{2}2s^2$

Thus $B$ need less energy to remove the outermost electron

Hence I.E of $B<$ I.E of $Be$

Atomic no. of $O=8\to 1s^{2}2s^{2}2p^4$

Atomic no. of $N=7\to 1s^{2}2s^{2}2p^3$

In case of $N$ its $p-$orbital is half filled which is more stable than other configuration, thus it required more energy to remove the outermost electron it become more stab;e with configuration

$\to 1s^{2}2s^{2}2p^3$

Thus it required less energy to remove the outermost electron and hence

I.E of O$<$ I.E of M

and because of high electronegativity of $F$, it is also difficult to remove the outermost electron of $F$.

Therefore , I.E of $F$$>$ I.E of $O$