Step 1: Analyze each ion.
- (A) Ni$^{2+}$: Configuration = [Ar] 3d$^8$ $\Rightarrow$ partially filled d-orbitals $\Rightarrow$ coloured.
- (B) Fe$^{3+}$: Configuration = [Ar] 3d$^5$ $\Rightarrow$ half-filled d-orbitals $\Rightarrow$ coloured.
- (C) Cu$^{2+}$: Configuration = [Ar] 3d$^9$ $\Rightarrow$ partially filled d-orbitals $\Rightarrow$ coloured.
- (D) Cu$^{+1}$: Configuration = [Ar] 3d$^{10}$ $\Rightarrow$ completely filled d-orbitals, no d–d transitions possible $\Rightarrow$ colourless.
Step 2: Final Answer.
Hence, the colourless ion is Cu$^{+1}$.
\[
\boxed{\text{Cu}^{+1}}
\]
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