Question

which set correctly orders the atoms from lowest to highest ionization energy? rb < na < si < s s < si < na < rb s < na < si < rb rb < na < s < si

Explanation:

Step1: Understand the trend of ionization energy

Ionization energy increases across a period (from left to right) and decreases down a group.

Step2: Analyze the positions of the elements

• \(Rb\) is in Group 1, Period 5.
• \(Na\) is in Group 1, Period 3. Since \(Rb\) is below \(Na\) in the same group, \(Rb\) has lower ionization energy than \(Na\) (\(Rb
• \(Si\) is in Group 14, Period 3 and \(S\) is in Group 16, Period 3. Since they are in the same period and \(S\) is to the right of \(Si\), \(Si < S\).
• Combining these, we get \(Rb

Answer:

\(Rb < Na < Si < S\) (the first option)