Step 1: Understanding the Question:
Transition metal ions exhibit color if they have unpaired electrons in their \( d \)-orbitals, which allows for d-d electron transitions by absorbing visible light.
Step 2: Key Formula or Approach:
Write the electronic configuration for each given ion and check for the presence of unpaired \( d \)-electrons.
Step 3: Detailed Explanation:
(A) \( \text{Sc}^{3+} \): Scandium (\( Z = 21 \)) is \( [\text{Ar}] \text{3d}^1 \text{4s}^2 \). The \( \text{Sc}^{3+} \) ion loses 3 electrons, becoming \( [\text{Ar}] \text{3d}^0 \). Since there are no \( d \)-electrons, it is colorless.
(B) \( \text{Ti}^{4+} \): Titanium (\( Z = 22 \)) is \( [\text{Ar}] \text{3d}^2 \text{4s}^2 \). The \( \text{Ti}^{4+} \) ion loses 4 electrons, becoming \( [\text{Ar}] \text{3d}^0 \). It is also colorless.
(C) \( \text{Zn}^{2+} \): Zinc (\( Z = 30 \)) is \( [\text{Ar}] \text{3d}^{10} \text{4s}^2 \). The \( \text{Zn}^{2+} \) ion loses 2 electrons, becoming \( [\text{Ar}] \text{3d}^{10} \). Because its \( d \)-orbitals are completely filled, no d-d transitions can occur, making it colorless.
(D) \( \text{Cr}^{3+} \): Chromium (\( Z = 24 \)) is \( [\text{Ar}] \text{3d}^5 \text{4s}^1 \). The \( \text{Cr}^{3+} \) ion loses 3 electrons, becoming \( [\text{Ar}] \text{3d}^3 \). It has 3 unpaired electrons in its \( d \)-orbital, which allows d-d transitions. Therefore, it forms colored compounds.
Step 4: Final Answer:
\( \text{Cr}^{3+} \) is the only ion with unpaired \( d \)-electrons and hence forms colored compounds.