Chemistry

Question Analysis
This question involves ordering a mix of neutral atoms (P, S, Cl) and an anion (Cl⁻) by atomic/ionic radius.
The main focus is on integrating multiple atomic radius trends (period trend, atom vs. anion).

Key Concept Explanation
Period trend (neutral atoms): Atomic radius decreases from left to right → P > S > Cl.
Atom vs. anion: Anions are larger than their parent atoms due to increased electron-electron repulsion → Cl⁻ > Cl.
Combined trend: Cl⁻ is larger than Cl, and P is the largest neutral atom in the group.

Step-by-Step Solution
1. Order the neutral atoms using the period trend: P (Period 3, Group 15) > S (Group 16) > Cl (Group 17).
2. Compare Cl⁻ to Cl: Cl⁻ > Cl.
3. Integrate the two orderings:
Cl⁻ is larger than Cl, but we need to compare Cl⁻ to P and S.
Cl⁻ has the same electron configuration as Ar (n = 3) and a lower nuclear charge than Cl.
P (n = 3, 15 protons) has a lower nuclear charge than Cl⁻ (17 protons), but Cl⁻ has more ...