Predict which of the following will be colored in its aqueous solution?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(Mn^{2+}\)
A transition metal ion is coloured when it has a partly filled d subshell. An electron in a lower energy d orbital absorbs light of a visible frequency and is excited to a higher energy d orbital (a d–d transition), and we see the complementary colour of the light absorbed. Ions with an empty (d0) or completely filled (d10) d subshell have no such transition and are colourless.
Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Mn^2+} \): \( \mathrm{3d^{5}} \). Its 3d subshell is partly filled, so \( \ce{Mn^2+} \) is coloured; its aqueous solution is pale pink.
Why the other options are incorrect
Option (1) \(Ti^{4+}\) — Incorrect. Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); losing all four outer electrons gives \( \ce{Ti^4+} \): \( \mathrm{3d^{0}} \), so it is colourless.
Option (2) \(Sc^{3+}\) — Incorrect. Sc (Z = 21) is \( \mathrm{[Ar]\,3d^{1}\,4s^{2}} \); losing three electrons gives \( \ce{Sc^3+} \): \( \mathrm{3d^{0}} \), colourless.
Option (4) \(Zn^{2+}\) — Incorrect. Zn (Z = 30) is \( \mathrm{[Ar]\,3d^{10}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Zn^2+} \): \( \mathrm{3d^{10}} \), a completely filled subshell, colourless.
NCERT (2026-27 edition):
Formation of Coloured Ions – …
