The colour of transition metal ions is typically due to the presence of unpaired $d$-electrons, which undergo $d-d$ transitions by absorbing light from the visible region. Let us examine the electronic configuration of each ion:

• $\text{Cr}$ ($Z=24$): $[\text{Ar}] 3d^5 4s^1 \Rightarrow \text{Cr}^{4+}$: $[\text{Ar}] 3d^2$ (2 unpaired electrons, so it is coloured)
• $\text{Sc}$ ($Z=21$): $[\text{Ar}] 3d^1 4s^2 \Rightarrow \text{Sc}^{3+}$: $[\text{Ar}] 3d^0$ (0 unpaired electrons, no $d-d$ transition possible, so it is colourless)
• $\text{Ti}$ ($Z=22$): $[\text{Ar}] 3d^2 4s^2 \Rightarrow \text{Ti}^{3+}$: $[\text{Ar}] 3d^1$ (1 unpaired electron, so it is coloured)
• $\text{V}$ ($Z=23$): $[\text{Ar}] 3d^3 4s^2 \Rightarrow \text{V}^{3+}$: $[\text{Ar}] 3d^2$ (2 unpaired electrons, so it is coloured) Thus, $\text{Sc}^{3+}$ is the colourless ion among the given options.