Final answer:
The ionization energy of boron is lower than that of beryllium, despite boron having a greater nuclear charge. This is due to beryllium having its outermost electron in an s subshell, which is more tightly bound, while boron's outermost electron is in a p subshell, which is higher in energy and easier to remove. Subshell introduction leads to these deviations from the general ionization energy trend.
Step-by-step explanation:
The characteristics of ionization energy exhibit specific trends within the periodic table. Ionization energy generally increases across a period due to the increased nuclear charge (Z) that causes a stronger attraction between the nucleus and the electrons, meaning more energy is required to remove an electron. However, there are deviations from this trend, particularly seen when comparing boron (Z=5) and beryllium (Z=4). Boron has a lower ionization energy than beryllium despite having a greater nuclear charge. This apparent contradiction can be explained by electronic configurations and subshell energies.
Beryllium has an electronic configuration of [He]2s², and thus the electron removed during ionization is an s electron which is lower in energy and more closely bound to the nucleus, making it harder to remove. In contrast, boron's configuration is [He]2s²2p¹, and the electron removed is a p electron which is higher in energy and easier to remove due to less penetration and more shielding effects. Therefore, a new subshell introduction, like the 2p in boron, causes a small deviation from the trend and a lower ionization energy.