Stable ions in Period 3 achieve noble gas electron configurations Steps: - Y in Period 3 (Z=11–18) forms stable ion with 18 electrons, so Y is P (Z=15, P³⁻), S (Z=16, S²⁻), or Cl (Z=17, Cl⁻), achieving Ar configuration. - X has lower Z than Y, so X is a metal (Na Z=11, Mg Z=12, or Al Z=13) forming cation Na⁺, Mg²⁺, or Al³⁺ with 10 electrons (Ne configuration). - X ion (10 electrons) has smaller radius than Y ion (18 electrons) because cations have higher effective nuclear charge and fewer electrons, pulling electrons closer. - The condition Y ion radius > X ion radius holds, confirming X ion has 10 electrons. Why A is correct: - A correctly assigns 10 electrons to X's stable ion, per octet rule for Period 3 metals achieving Ne configuration. Why the others are wrong: - B: Assigns 18 electrons to Y, but this describes the larger ion, not the key deduction for X. - C: Assigns 10 electrons to Y, but Y (nonmetal) forms anion with 18 electrons. - D: Assigns 18 electrons …