Question

Rank the elements \(\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{N}\), and \(\mathrm{O}\) in order of increasing first ionization energy. Explain your reasoning.

Short answer

The elements $\mathrm{Be}$, $\mathrm{B}$, $\mathrm{C}$, $\mathrm{N}$, and $\mathrm{O}$ can be ranked in order of increasing first ionization energy as follows: \(\mathrm{Be} < \mathrm{B} < \mathrm{C} < \mathrm{N} < \mathrm{O}\). This is because ionization energy increases across a period from left to right in the periodic table, and all of these elements are in the second period. Be has the lowest ionization energy, while O has the highest ionization energy among the given elements.

Step-by-step solution

1. Determine the positions of the given elements in the periodic table

To predict the trend in ionization energy, we first need to determine the positions of Be, B, C, N, and O in the periodic table. These elements are all in the second period, with Be being an alkaline earth metal (Group 2) and O being a chalcogen (Group 16). B, C, and N belong to groups 13, 14, and 15, respectively.

2. Apply the periodic trend of ionization energy

We know that ionization energy increases across a period from left to right. Therefore, as we move from Be to O along the second period in the periodic table, the ionization energy increases. So, Be will have the lowest ionization energy, and O will have the highest ionization energy out of these elements. B, C, and N will have increasing ionization energies as we move from left to right.

3. Rank the elements in order of increasing first ionization energy

Based on the periodic trend of ionization energy, we can rank the elements in order of increasing first ionization energy as follows: Be < B < C < N < O. Be has the lowest ionization energy, while O has the highest ionization energy among the given elements.