Match the ions given in List-I with number of unpaired electrons given in List-II
|
List-I Ion |
List-II Number of unpaired electrons |
|
(A) $V^{2+}$ |
(I) 1 |
|
(B) $Cr^{2+}$ |
(II) 4 |
|
(C) $Ti^{2+}$ |
(III) 3 |
|
(D) $Cu^{2+}$ |
(IV) 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
|
List-I Ion |
List-II Number of unpaired electrons |
|
(A) $V^{2+}$ |
(III) 3 |
|
(B) $Cr^{2+}$ |
(II) 4 |
|
(C) $Ti^{2+}$ |
(IV) 2 |
|
(D) $Cu^{2+}$ |
(I) 1 |
(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