Rank the following elements in order of increasing ionization energy: Br, F, Ga, K and Se.

• A. Ga < K < Br < Se < F
• B. Ga < Br < K < Se < F
• C. K < Se < Ga < Br < F
• D. K < Ga < Se < Br < F
• E. K < Ga < Se < F < Br

Answer: (D)

Verified step by step guidance

1. Understand the concept of ionization energy: Ionization energy is the energy required to remove an electron from an atom in the gaseous state. Generally, ionization energy increases across a period and decreases down a group in the periodic table.
2. Identify the position of each element in the periodic table: Br (Bromine), F (Fluorine), Ga (Gallium), K (Potassium), and Se (Selenium). Fluorine and Bromine are halogens, Gallium is a post-transition metal, Potassium is an alkali metal, and Selenium is a non-metal.
3. Consider the periodic trends: As you move from left to right across a period, ionization energy increases. As you move down a group, ionization energy decreases. Therefore, elements further to the right and higher up in the periodic table generally have higher ionization energies.
4. Apply the trends to the given elements: Potassium (K) is in Group 1 and has the lowest ionization energy. Gallium (Ga) is further right than Potassium but lower than Selenium (Se), Bromine (Br), and Fluorine (F). Fluorine, being the furthest right and highest up, has the highest ionization energy.
5. Rank the elements based on the trends: Start with the element with the lowest ionization energy (K), followed by Ga, Se, Br, and finally F, which has the highest ionization energy.