Which of these ions is expected to be colored in aqeous solution?
[I] `Fe^(3+)` [II] `Zn^(2+)` [III] `AI^(3+)` [IV] `Sc^(3+)`

Options

A. [I] only
B. [III] only
C. [I] and [II] only
D. [I], [II] and [III]

Answer

To determine which of the given ions is expected to be colored in aqueous solution, we need to analyze the electronic configurations of each ion and check for the presence of unpaired electrons. The presence of unpaired electrons in d-orbitals is what typically leads to the coloration of transition metal ions.

### Step-by-Step Solution:

1. **Analyze Fe^(3+)**:
- The electronic configuration of Fe (Iron) is:
- Atomic number = 26
- Configuration: [Ar] 4s² 3d⁶
- For Fe^(3+), we remove 3 electrons (2 from 4s and 1 from 3d):
- Fe^(3+) configuration: [Ar] 3d⁵
- In 3d⁵, there are 5 unpaired electrons.
- **Conclusion**: Fe^(3+) is colored due to the presence of unpaired electrons.

2. **Analyze Zn^(2+)**:
- The electronic configuration of Zn (Zinc) is:
- Atomic number = 30
- Configuration: [Ar] 4s² 3d¹⁰
- For Zn^(2+), we remove 2 electrons (both from 4s):
- Zn^(2+) configuration: [Ar] 3d¹⁰
- In 3d¹⁰, there are no unpaired electrons.
- **Conclusion**: Zn^(2+) is not colored.

3. **Analyze Al^(3+)**:
- The electronic configuration of Al (Aluminum) is:
- Atomic number = 13
- Configuration: [Ne] 3s² 3p¹
- For Al^(3+), we remove 3 electrons (2 from 3s and 1 from 3p):
- Al^(3+) configuration: [Ne]
- In [Ne], there are no unpaired electrons.
- **Conclusion**: Al^(3+) is not colored.

4. **Analyze Sc^(3+)**:
- The electronic configuration of Sc (Scandium) is:
- Atomic number = 21
- Configuration: [Ar] 4s² 3d¹
- For Sc^(3+), we remove 3 electrons (2 from 4s and 1 from 3d):
- Sc^(3+) configuration: [Ar] 3d⁰
- In 3d⁰, there are no unpaired electrons.
- **Conclusion**: Sc^(3+) is not colored.

### Final Conclusion:
- The only ion that is expected to be colored in aqueous solution is **Fe^(3+)**. Therefore, the answer is **[I] Fe^(3+)**.