Arrange the following transition metal ions in the increasing order of unpaired electrons.
(A) \( Cu^{2+} \)
(B) \( Cr^{2+} \)
(C) \( Ni^{2+} \)
(D) \( Sc^{3+} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (D), (A), (C), (B)
The number of unpaired electrons is read from the d electron count of the ion. Write the configuration of the neutral atom from its atomic number, remove the 4s electrons first and then 3d electrons, and fill the five 3d orbitals singly before pairing (Hund's rule).
(A) \( \ce{Cu^2+} \): Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); losing the 4s and one 3d electron gives \( \mathrm{3d^{9}} \): 1 unpaired electron.
(B) \( \ce{Cr^2+} \): Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \); losing the 4s and one 3d electron gives \( \mathrm{3d^{4}} \): 4 unpaired electrons.
(C) \( \ce{Ni^2+} \): Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{8}} \): 2 unpaired electrons.
(D) \( \ce{Sc^3+} \): Sc (Z = 21) is \( \mathrm{[Ar]\,3d^{1}\,4s^{2}} \); losing all three gives \( \mathrm{3d^{0}} \): 0 unpaired electrons.
Shortcut: \( \ce{Sc^3+} \) has no d electrons and \( \ce{Cr^2+} \) the most unpaired ones. Only Option (4) starts with (D) and ends with (B).