Which of the following ion is likely to form white complexes

Options

A. `Cd^(2+)`
B. `Ti^(3+)`
C. `Cr^(3+)`
D. none of the above

Answer

To determine which ion is likely to form white complexes, we need to analyze the electronic configurations of the given ions and their ability to undergo d-d transitions.

### Step-by-Step Solution:

1. **Identify the Ions**: The ions in question are Cadmium (Cd²⁺), Titanium (Ti³⁺), and Chromium (Cr³⁺).

2. **Determine the Electronic Configuration**:
- **Cadmium (Cd)**: The atomic number of cadmium is 48. The electronic configuration is:
- Neutral Cd: [Kr] 4d¹⁰ 5s²
- For Cd²⁺: Remove 2 electrons from the outermost shell (5s), resulting in [Kr] 4d¹⁰.
- **Titanium (Ti)**: The atomic number of titanium is 22. The electronic configuration is:
- Neutral Ti: [Ar] 4s² 3d²
- For Ti³⁺: Remove 3 electrons (2 from 4s and 1 from 3d), resulting in 3d¹.
- **Chromium (Cr)**: The atomic number of chromium is 24. The electronic configuration is:
- Neutral Cr: [Ar] 4s² 3d⁵
- For Cr³⁺: Remove 3 electrons (2 from 4s and 1 from 3d), resulting in 3d³.

3. **Analyze the d-Orbital Electrons**:
- **Cd²⁺**: [Kr] 4d¹⁰ has all paired electrons (10 electrons in d-orbitals). Therefore, it cannot undergo d-d transitions.
- **Ti³⁺**: 3d¹ has 1 unpaired electron. It can undergo d-d transitions, which would result in color.
- **Cr³⁺**: 3d³ has 3 unpaired electrons. It can also undergo d-d transitions, which would result in color.

4. **Conclusion**: Since Cd²⁺ has all paired electrons and cannot undergo d-d transitions, it is likely to form a white complex. The other ions (Ti³⁺ and Cr³⁺) have unpaired electrons and can exhibit color due to d-d transitions.

### Final Answer:
**Cadmium ion (Cd²⁺) is likely to form white complexes.**

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