Step 1: Colour in transition-metal ions arises from d-d transitions, which are only possible when the d subshell is partially filled (has unpaired electrons and empty d orbitals to jump into).
Step 2: Work out the d-electron configurations:
\( Cu^{+} \): \(3d^{10}\) (completely filled, no d-d transition possible) \(\rightarrow\) colourless.
\( Cu^{2+} \): \(3d^{9}\) (one unpaired electron) \(\rightarrow\) blue.
\( Ni^{2+} \): \(3d^{8}\) (unpaired electrons) \(\rightarrow\) green.
\( Co^{2+} \): \(3d^{7}\) (unpaired electrons) \(\rightarrow\) pink.
Step 3: Only \( Cu^{+} \) has a fully filled \(3d^{10}\) configuration, so no d-d transition and no colour.
Correct answer: (i) \( Cu^{+} \) ion.