Question:Figure 8.16 shows that in the Z = 3 to 10 filling of the n = 2 shell (lithium to neon), there is an upward trend in elements' first Ionization energies. Why is there a drop as Z goes from 4 to 5, from beryllium to boron?

It is easier with lesser ionization energy to remove the unpaired electron in the subshell2p¹ of Boron (B)than the electron in the closed subshellof Beryllium (Be)in the process of ionization energy from to L_i(z = 3 to 10).

Beryllium has an electronic configuration . The subshells are closed. Electrons in the closed shells are very tightly bound and the positive nuclear charge is very large when compared to the negative charge of the inner shielding electrons. So, these electrons cannot be easily detached. So, higher energy is needed to remove these electrons, whereas, Boron has the electronic configuration $1s^{2}2s^{2}2p^1$. As the subshell is not closed, electrons in this shell could be removed with laser ionization energy compared to Beryllium.