Question

which of the following elements has the lowest ionization energy?
he
f
rb
c
na

Explanation:

Ionization energy is the energy required to remove an electron from an atom. In the periodic table, ionization energy generally increases across a period (from left to right) and decreases down a group.

• He (Helium): It is a noble gas with a full valence shell (\(1s^{2}\)). Noble gases have very high ionization energies because of their stable electron configuration.
• F (Fluorine): It is a non - metal in Group 17. Non - metals (except noble gases) have relatively high ionization energies as they tend to gain electrons rather than lose them. Fluorine has a high effective nuclear charge, which holds its electrons tightly.
• Rb (Rubidium): It is an alkali metal in Group 1. Alkali metals have low ionization energies. Rubidium is in the fifth period. As we move down Group 1 (from Li to Cs), the atomic radius increases, and the outermost electron is further from the nucleus, so less energy is required to remove it.
• C (Carbon): It is a non - metal in Group 14. Carbon has a higher ionization energy than alkali metals. It has a smaller atomic radius compared to alkali metals (same period as some alkali metals in lower periods), and its electrons are more tightly bound.
• Na (Sodium): It is an alkali metal in Group 1, but it is in the third period. Rubidium (in the fifth period) has a larger atomic radius than sodium. Since the outermost electron in Rb is further from the nucleus (due to more electron shells), it is easier to remove compared to the outermost electron in Na.

Answer:

Rb