Which of the following arrangements indicates the correct decreasing order of ionization energy of period 3 elements?

Analyze the question: You're asked to arrange the period 3 elements Na, Al, P, and Cl in order of decreasing ionization energy, from highest to lowest.

Concept: Across a period, from left to right:

• Nuclear charge increases.
• Atomic radius decreases, because electrons are pulled in more strongly while the shell stays the same.
• It becomes harder to remove an outer electron, so first ionization energy generally increases across the period.

There are small exceptions (for example, Al is lower than Mg, and S is lower than P), but they don't change the order of these four elements.

Atomic numbers (period 3 order): Na (11) → Al (13) → P (15) → Cl (17)

Approximate first ionization energies (kJ/mol):

• Cl: ~1251
• P: ~1012
• Al: ~578
• Na: ~496

Decreasing order: Cl > P > Al > Na

Checking the options:

• A. Na, Al, P, Cl is the increasing order, so it is wrong.
• B. Cl, P, Al, Na matches the decreasing order, so it is correct.
• C. Na, P, Cl, Al is not in any consistent order, so it is wrong.
• D. Al, Na, P, Cl is not in any consistent order, so it is wrong.

**Answer: B.

Tip: Ionization energy increases across a period and decreases down a group. For "decreasing order" questions, start with the element furthest to the right, since that is usually the highest. Here, that is Cl, and only option B starts with it.