Final answer:
Boron displays a sudden increase in its ionization energy when the fourth electron is removed, transitioning from the valence shell to the core 1s² subshell, which requires substantially more energy than the removal of the first three electrons.
Step-by-step explanation:
The sudden increase in ionization energy for boron would occur when removing the fourth electron. This is because the first three electrons removed during the ionization of boron are from the 2s and 2p subshells. However, removing the fourth electron means removing an electron from the filled 1s² subshell, which is closer to the nucleus and more strongly bound due to the lower energy and increased penetration of s orbitals. Therefore, this step requires considerably more energy.
According to the given information, for boron, the ionization energy dramatically increases from the third to the fourth ionization energy because the fourth ionization means removing an electron from the noble gas core configuration of helium ([He]), resulting in a fully ionized B4+ ion with a noble gas configuration.
It is important to note that ionization energies typically increase progressively as you remove electrons from an atom due to the decreasing electron-electron repulsion and relatively stronger nuclear attraction on the remaining electrons.