Step 1: Rule of thumb: a transition-metal ion is coloured only if it has a partly filled d subshell; a \(d^0\) or \(d^{10}\) ion is colourless.
Step 2: Copper is atomic number 29, so \(Cu^{+}\) loses one electron beyond the \(Cu\) configuration and ends up as \(3d^{10}\), a full subshell with zero unpaired electrons.
Step 3: With every d orbital already full there is no vacant d level for an electron to be promoted into, so visible light is not absorbed and the ion looks colourless.
Step 4: The other three ions (\(Cu^{2+}=3d^9\), \(Ni^{2+}=3d^8\), \(Co^{2+}=3d^7\)) all keep unpaired d electrons and therefore absorb visible light, appearing blue, green and pink respectively.
\[\boxed{Cu^{+}\ (3d^{10})\ \text{is colourless}}\]