Arrange the following ions in increasing order of number of 3d-electrons
(A) \( Cr^{2+} \)
(B) \( Cu^+ \)
(C) \( Ti^{3+} \)
(D) \( Mn^+ \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C), (A), (D), (B)
To count the 3d electrons of a transition metal ion, first write the configuration of the neutral atom from its atomic number, then remove the electrons lost in forming the ion. The 4s electrons are removed first, and only then the 3d electrons.
(A) \( \ce{Cr^2+} \): Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \). Losing the 4s electron and one 3d electron gives \( \mathrm{[Ar]\,3d^{4}} \): 4 electrons.
(B) \( \ce{Cu+} \): Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \). Losing the 4s electron gives \( \mathrm{[Ar]\,3d^{10}} \): 10 electrons.
(C) \( \ce{Ti^3+} \): Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \). Losing both 4s electrons and one 3d electron gives \( \mathrm{[Ar]\,3d^{1}} \): 1 electron.
(D) \( \ce{Mn+} \): Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \). Losing one 4s electron gives \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \): 5 electrons.
In increasing order of 3d electrons:
\( \displaystyle \ce{Ti^3+}\,(1) \lt \ce{Cr^2+}\,(4) \lt \ce{Mn+}\,(5) \lt \ce{Cu+}\,(10) \)
Shortcut: \( \ce{Ti^3+} \) has the fewest 3d electrons and \( \ce{Cu+} \) a full 3d set, so the order starts with (C) and ends with (B): only Options (2) and (3). Since \( \ce{Cr^2+} \) has fewer 3d electrons than \( \ce{Mn+} \), (A) comes before (D), which leaves Option (2).