Which of the following ions has an electron configuration different from the other?

Step 1: Analyze the question We need to find the ion whose electron configuration is different from the others. So we count the electrons in each ion and compare.

Step 2: Concept

• Anion (negative): electrons = atomic number + charge
• Cation (positive): electrons = atomic number − charge

Step 3: Work through each option

| Option | Ion | Electron count | Configuration | |---|---|---|---| | A | Cl⁻ (Z = 17) | 17 + 1 = 18 | 2, 8, 8 | | B | O²⁻ (Z = 8) | 8 + 2 = 10 | 2, 8 | | C | Mg²⁺ (Z = 12) | 12 − 2 = 10 | 2, 8 | | D | Al³⁺ (Z = 13) | 13 − 3 = 10 | 2, 8 |

Step 4: Compare O²⁻, Mg²⁺ and Al³⁺ are all isoelectronic with neon (2, 8). Cl⁻ is the odd one out, with 18 electrons (2, 8, 8), which is the argon configuration.

Answer: A. Cl⁻

Exam tip: Quickly convert every ion to its electron count. Ions from Period 2 and Period 3 metals (Na⁺, Mg²⁺, Al³⁺) and nonmetals (N³⁻, O²⁻, F⁻) all have the neon configuration (10 electrons). Chlorine, sulfur and phosphorus ions (Cl⁻, S²⁻, P³⁻), along with K⁺ and Ca²⁺, have the argon configuration (18 electrons).