COMEDK 2025 Afternoon Solved Paper

Concept:Determine the number of unpaired electrons in each ion using electronic configuration and Hund’s rule.Explanation:For each ion, write the d‑electron count after removal of electrons.• Cr²⁺ (Z = 24): Neutral Cr is $[\mathrm{Ar}]\,3d^{5}4s^{1}$. Cr²⁺ loses the 4s electron and one 3d electron → $3d^{4}$. Four electrons occupy four orbitals singly → 4 unpaired electrons.• Fe³⁺ (Z = 26): Neutral Fe is $[\mathrm{Ar}]\,3d^{6}4s^{2}$. Fe³⁺ loses two 4s and one 3d electron → $3d^{5}$. Five electrons each in separate orbitals → 5 unpaired electrons.• Ni²⁺ (Z = 28): Neutral Ni is $[\mathrm{Ar}]\,3d^{8}4s^{2}$. Ni²⁺ removes the two 4s electrons → $3d^{8}$. Following Hund’s rule, 5 orbitals get 5 singly, remaining 3 pair up → 2 unpaired electrons.• Cu²⁺ (Z = 29): Neutral Cu is $[\mathrm{Ar}]\,3d^{10}4s^{1}$. Cu²⁺ loses the 4s and one 3d electron → $3d^{9}$. Nine electrons fill 4 orbitals completely and leave one orbital singly occupied → 1 unpaired electron.Summarising unpaired electrons:Cu²⁺ (B): 1, Ni²⁺ (C): 2, Cr²⁺ (A): 4, Fe³⁺ (D): 5.Hence increasing order: B < C < A < D.Answer:Option A: B < C < A < D.