Chemistry

Question Analysis
This question involves comparing ionization energies across both a period and a group. The main focus is on applying multiple trend factors (atomic radius, shielding effect) to rank IE values for three specific elements.

Key Concept Explanation
Across Period 2 (Li to Be): IE increases (smaller radius, higher nuclear charge).
Down Group 1 (Li to Na): IE decreases (larger radius, more inner electron shells leading to greater shielding).
Sodium is in a lower group than Li and Be, so it has a larger radius and stronger shielding effect, making it easier to remove a valence electron—resulting in the lowest IE among the three.

Step-by-Step Solution
1. Apply period trend: Li < Be (IE increases left to right).
2. Apply group trend: Li > Na (IE decreases top to bottom).
3. Combine trends to rank IE: Na < Li < Be. The element with the lowest IE is sodium (Na).
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