Rank the elements Be, B, C, N, and O in order of increasing first ionization energy. Explain your reasoning.

First ionization energy is the energy required to remove the most loosely bound electron from an atom in its ground state. Generally, ionization energy increases across a period from left to right and decreases down a group.

Write the electronic configuration of the given elements: Be (Beryllium): \(1s^2 2s^2\) B (Boron): \(1s^2 2s^2 2p^1\) C (Carbon): \(1s^2 2s^2 2p^2\) N (Nitrogen): \(1s^2 2s^2 2p^3\) O (Oxygen): \(1s^2 2s^2 2p^4\)

Ionization energy typically increases across a period due to the increased effective nuclear charge, which attracts the outer electrons more strongly. As you move from left to right, the atomic number increases, and there is a greater attraction between the positively charged nucleus and the negatively charged electrons, making it harder to remove an electron.

The shielding effect is when inner electrons shield or partially block the nuclear charge (attraction) felt by the outer electrons. In some cases, this can cause the ionization energy to deviate from the general trend. For instance, there is a significant increase in ionization energy between B (Boron) and C (Carbon) due to the fact that the electron removed from Carbon is from a filled 2s subshell, which has a slightly higher ionization energy compared to the 2p subshell. Similarly, there is a slight decrease in ionization energy from N (Nitrogen) to O (Oxygen). Nitrogen has a half-filled 2p subshell, which makes it more stable and require more energy to remove an electron. In Oxygen, due to the repulsion between electrons in the same orbital, the ionization energy gets slightly reduced.

Taking into account the electronic configurations and factors that impact ionization energy, we can rank the elements in order of increasing first ionization energy: Be < B < C < N > O The final ranking is Be, B, O, C, N.