The set having ions which are coloured and paramagnetic both is -

For a transition-metal ion to be both coloured and paramagnetic, it must have unpaired d-electrons. Ions with a $$d^0$$ or $$d^{10}$$ configuration are generally colourless and diamagnetic.

Option (1): $$\text{Cu}^{2+}$$ has the configuration $$[\text{Ar}]\,3d^9$$ — one unpaired electron, so it is paramagnetic and coloured (blue/green). $$\text{Cr}^{3+}$$ has $$[\text{Ar}]\,3d^3$$ — three unpaired electrons, paramagnetic and coloured (green/violet). $$\text{Sc}^{+}$$ has $$[\text{Ar}]\,3d^1\,4s^1$$ — unpaired electrons present, so it is paramagnetic and can exhibit colour due to partially filled d-orbitals. All three ions in this set satisfy both conditions.

Option (2): $$\text{Zn}^{2+}$$ is $$3d^{10}$$ — fully filled, diamagnetic and colourless. This set fails.

Option (3): $$\text{Sc}^{3+}$$ is $$3d^0$$, $$\text{V}^{5+}$$ is $$3d^0$$, $$\text{Ti}^{4+}$$ is $$3d^0$$ — all are diamagnetic and colourless. This set fails.

Option (4): $$\text{Mn}^{7+}$$ is $$3d^0$$ (diamagnetic, colourless) and $$\text{Hg}^{2+}$$ is $$5d^{10}$$ (diamagnetic, colourless). This set fails.

The answer is option (1).