Match the ions given in List-I with number of unpaired electrons given in List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(II), (C)-(IV), (D)-(III) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) |
(A)-(III), (B)-(II), (C)-(IV), (D)-(I) |
The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(A) V²⁺: Atomic number of V is 23 → [Ar] 4s² 3d³. V²⁺ loses 2 electrons → 3d³ (high-spin in typical cases). 3 unpaired electrons in d orbitals. → (III) 3 (B) Cr²⁺: Atomic number of Cr is 24 → [Ar] 4s¹ 3d⁵. Cr²⁺ → 3d⁴. 4 unpaired electrons. → (II) 4 (C) Ti²⁺: Atomic number of Ti is 22 → [Ar] 4s² 3d². Ti²⁺ → 3d². 2 unpaired electrons. → (IV) 2 (D) Cu²⁺: Atomic number of Cu is 29 → [Ar] 4s¹ 3d¹⁰. Cu²⁺ → 3d⁹. 1 unpaired electron (due to one hole in d orbitals). → (I) 1 |