Which of the following elements requires the most energy to lose one electron (i.e., has the highest first ionization energy)?

• A. Neon (Ne)
• B. Aluminum (Al)
• C. Sodium (Na)
• D. Potassium (K)

Answer: (A)

Verified step by step guidance

1. Understand that the first ionization energy is the energy required to remove one electron from a neutral atom in the gaseous state.
2. Recall the general trend of ionization energy in the periodic table: it increases across a period (left to right) and decreases down a group (top to bottom).
3. Identify the positions of the given elements in the periodic table: Potassium (K), Sodium (Na), Aluminum (Al), and Neon (Ne). Potassium and Sodium are in Group 1, Aluminum is in Group 13, and Neon is a noble gas in Group 18.
4. Apply the trend: since Neon is at the far right of its period and is a noble gas with a full valence shell, it has a much higher ionization energy compared to the metals listed, which are more willing to lose electrons.
5. Conclude that Neon requires the most energy to remove one electron because it has a stable electron configuration and is located at the top right of the periodic table among the given options.