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Which of the following ions will exhibit colour in aqueous solutions?
A
$Sc^{3+}$ (Z = 21)
B
$La^{3+}$ (Z = 57)
C
$Ti^{3+}$ (Z = 22)
D
$Lu^{3+}$ (Z = 71)
Detailed Solution
An ion is coloured in aqueous solution when it has unpaired electrons in d (or f) orbitals that can undergo d–d (or f–f) transitions by absorbing visible light.
$Sc^{3+}$: $[Ar]3d^0$, no d electrons, colourless.
$La^{3+}$: $[Xe]4f^0$, no f electrons, colourless.
$Lu^{3+}$: $[Xe]4f^{14}$, completely filled, no unpaired electron, colourless.
$Ti^{3+}$: $[Ar]3d^1$, one unpaired d electron; the $t_{2g} \rightarrow e_g$ transition absorbs visible light, so $[Ti(H_2O)_6]^{3+}$ is purple.
Hence $Ti^{3+}$ exhibits colour.
$Sc^{3+}$: $[Ar]3d^0$, no d electrons, colourless.
$La^{3+}$: $[Xe]4f^0$, no f electrons, colourless.
$Lu^{3+}$: $[Xe]4f^{14}$, completely filled, no unpaired electron, colourless.
$Ti^{3+}$: $[Ar]3d^1$, one unpaired d electron; the $t_{2g} \rightarrow e_g$ transition absorbs visible light, so $[Ti(H_2O)_6]^{3+}$ is purple.
Hence $Ti^{3+}$ exhibits colour.