The following ions have the same electron configuration except  $_{8}$O, $_{12}$Mg, $_{13}$AI, $_{17}$CI

Analyze the question: We need to find the ion whose electron configuration is different from the others. Atoms form ions by gaining or losing electrons, so we count the electrons in each ion.

Concept: Ions with the same number of electrons are isoelectronic, so they have the same electron configuration.

Solution:

• Cl⁻: Z = 17, gains 1 electron, so 18 electrons: 2, 8, 8 (1s² 2s² 2p⁶ 3s² 3p⁶)
• O²⁻: Z = 8, gains 2 electrons, so 10 electrons: 2, 8 (1s² 2s² 2p⁶)
• Mg²⁺: Z = 12, loses 2 electrons, so 10 electrons: 2, 8
• Al³⁺: Z = 13, loses 3 electrons, so 10 electrons: 2, 8

O²⁻, Mg²⁺ and Al³⁺ all have 10 electrons, the same configuration as neon (1s² 2s² 2p⁶). Cl⁻ has 18 electrons, the same configuration as argon, so it is the odd one out.

Answer: A. Cl⁻.

Tip: For "same electron configuration" questions, always calculate the electron count: protons ± charge. Remember that anions come from Period 2 and 3 nonmetals gaining electrons, and metal cations lose electrons to reach the noble gas configuration of the preceding period (Ne for O²⁻, Mg²⁺, Al³⁺, and Ar for Cl⁻).