The increasing order of electron affinity of the electronic configurations of element is : (I) 1s2 2 — Periodic Table and Periodicity Chemistry Question

💡 Solution & Explanation

Step 1: Identify that (I) represents Chlorine (Cl), (II) represents Nitrogen (N), (III) represents Fluorine (F), and (IV) represents Sodium (Na). Step 2: Nitrogen has a stable, half-filled 2p3 subshell, giving it the lowest electron affinity. Sodium has a low electron affinity typical of alkali metals. Chlorine has a higher electron affinity than Fluorine because Chlorine's larger 3p subshell experiences less electron-electron repulsion than the compact 2p subshell of Fluorine. Step 3: Combining these trends gives the increasing order of electron affinity: II < IV < III < I, which matches option (a).