Final answer:
Boron has a smaller first ionization energy than beryllium because the electron removed during boron's ionization is a 2p electron, which is higher in energy and easier to remove compared to the 2s electron removed from beryllium. This is despite boron's greater nuclear charge.
Step-by-step explanation:
The question pertains to why boron has a smaller first ionization energy compared to beryllium. Ionization energy increases across a period as the effective nuclear charge (Zeff) increases. Boron (B), which has an atomic number of 5, should intuitively have a higher ionization energy than beryllium (Be), with an atomic number of 4, due to its greater nuclear charge. However, a deviation from this trend is observed because the type of electron being removed is from a different subshell.
Beryllium has a full 2s subshell ([He]2s2), which requires more energy to remove an electron from, compared to boron, which has its outermost electron in the 2p subshell ([He]2s22p1). Within the same shell, s electrons are held more tightly due to better penetration towards the nucleus and less shielding effect compared to p electrons. Therefore, although boron has a greater nuclear charge, its ionization energy is lower because the p electron in a higher energy level is easier to remove than the s electron in beryllium.