Identifying Period 3 element by ionization energy trend and oxide reactivity Steps: - Period 3 choices: Na (Z=11), Al (13), Si (14), S (16); element with one less proton means previous in sequence. - Ionization energy (IE) trend: IE(Na) < IE(Ne), IE(Al) < IE(Mg) due to p-electron removal; IE(Si) > IE(Al), IE(S) > IE(P). - Oxide reactivity: Na2O + H2O → 2NaOH (reacts); Al2O3 insoluble (no reaction); SiO2 insoluble (no reaction); SO2 + H2O → H2SO3 (reacts). - Only Al satisfies both lower IE than previous and non-reactive oxide. Why A is correct: - Al (Z=13) has first IE lower than Mg (Z=12) per periodic trend (s²p¹ configuration eases p-electron loss), and Al2O3 does not dissolve or react with water. Why the others are wrong: - B. Silicon: First IE higher than Al due to increased nuclear charge and stable half-filled p subshell. - C. Sodium: Na2O reacts vigorously with water to form alkaline NaOH solution. - D. Sulfur: First IE higher than P (stable half-filled p³ to p⁴), and SO2 reacts with water to form acidic sulfurous acid. Final …