Question

what is the predicted order of first ionization energies from highest to lowest for lithium (li), sodium (na), potassium (k), and rubidium (rb)?
○ rb > k > na > li
○ k > rb > na > li
○ li > na > k > rb
○ rb > k > li > na

Explanation:

Ionization energy is the energy required to remove an electron from an atom. For elements in the same group (like Li, Na, K, Rb which are in Group 1), as we move down the group, the atomic radius increases because there are more electron shells. The outermost electron is further from the nucleus and experiences less effective nuclear charge. So, it becomes easier to remove the outermost electron, meaning ionization energy decreases down the group.

Lithium (Li) is at the top of Group 1, then Sodium (Na), then Potassium (K), then Rubidium (Rb) as we go down. So the first ionization energy should be highest for Li, then Na, then K, then Rb.

Answer:

Li > Na > K > Rb