JEE Main 2026 Chemistry question
Match List-I with List-II. List-I (Electronic configuration of neutral atom (where n = 2)): A. ns², B. ns²np¹, C. ns²np³, D. ns²np⁶. List-II (1st Ionization Energy (kJ mol⁻¹)): I. 2080, II. 899, III. 800, IV. 1402.
Answer: (A)
Solution
With n=2: ns² is Be (899), ns²np¹ is B (800), ns²np³ is N (1402, half-filled stability), ns²np⁶ is Ne (2080).