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)
To count unpaired electrons, write the configuration of the neutral atom from its atomic number, remove the electrons lost in forming the ion (4s electrons first, then 3d), and fill the remaining 3d electrons singly into the five orbitals before any pairing (Hund’s rule).
(A) \( V^{2+} \) — This matches (III) 3. V (Z = 23) is \( \mathrm{[Ar]\,3d^{3}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{3}} \): three unpaired electrons.
(B) \( Cr^{2+} \) — This matches (II) 4. Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \); losing the 4s electron and one 3d electron gives \( \mathrm{3d^{4}} \): four unpaired electrons.
(C) \( Ti^{2+} \) — This matches (IV) 2. Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{2}} \): two unpaired electrons.
(D) \( Cu^{2+} \) — This matches (I) 1. Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); losing the 4s electron and one 3d electron gives \( \mathrm{3d^{9}} \): four pairs and one unpaired electron.
Shortcut: (A)-(III) appears only in Options (2) and (4), and (C)-(IV) then leaves only Option (4).