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IguaneCalme4548

i dont understand the ionization periodic trend. why is the comparaison point fluorine and not helium?

thank you

Mélissa S

Hi,

The comparison point for understanding the ionization energy periodic trend is typically chosen to be fluorine rather than helium because fluorine provides a more informative point for analysis due to its electron configuration. The ionization energy is the energy required to remove an electron from an atom. When you move across a period (a horizontal row) on the periodic table, the ionization energy generally increases.

Helium is an exception because it has only two electrons and is in the first period of the periodic table. It has a stable electron configuration (1s²) with a fully filled 1s orbital. Since helium is already at its lowest energy state (with a completely filled energy level), it has the highest ionization energy of any element in the periodic table. Removing an electron from helium would require breaking a stable electron configuration, and this requires a significant amount of energy.

Fluorine, on the other hand, is in the second period of the periodic table, and its electron configuration is 1s² 2s² 2p⁵. Fluorine is just one electron short of achieving a stable, fully filled 2p orbital. Because of this, it has a high tendency to gain an additional electron to achieve a stable configuration like neon (1s² 2s² 2p⁶). Removing an electron from fluorine requires less energy than removing one from helium, but it still requires a substantial amount of energy. This makes fluorine a suitable point for analyzing the trend across a period, as ionization energy generally increases from left to right across the periodic table due to the increasing nuclear charge and effective nuclear pull on the electrons.

So, by comparing fluorine to helium, you can observe how the ionization energy decreases as you move from right to left across a period, which is a fundamental aspect of the periodic trend.

Hope it helps! See you soon. :D