The ions from among the following which are colourless are `:`
`(i)Ti^(+4)" "(ii)Cu^(+1)" "(iii)Co^(+3)" "(iv)Fe^(+2)`

Explanation

To determine which of the given ions are colorless, we need to analyze their electronic configurations and the presence of unpaired electrons. Colorless ions typically have no unpaired electrons, as unpaired electrons can lead to d-d transitions that cause color.

### Step-by-Step Solution:

1. **Identify the Ions and Their Charges:**
- (i) Ti^(+4)
- (ii) Cu^(+1)
- (iii) Co^(+3)
- (iv) Fe^(+2)

2. **Determine the Electronic Configuration of Each Ion:**
- **(i) Ti^(+4)**:
- Titanium (Ti) has an atomic number of 22. Its ground state electronic configuration is [Ar] 3d² 4s².
- For Ti^(+4), we remove 4 electrons (2 from 4s and 2 from 3d), resulting in [Ar] 3d⁰.
- **Conclusion**: 0 unpaired electrons → Colorless.

- **(ii) Cu^(+1)**:
- Copper (Cu) has an atomic number of 29. Its ground state electronic configuration is [Ar] 3d¹⁰ 4s¹.
- For Cu^(+1), we remove 1 electron (from 4s), resulting in [Ar] 3d¹⁰.
- **Conclusion**: 0 unpaired electrons → Colorless.

- **(iii) Co^(+3)**:
- Cobalt (Co) has an atomic number of 27. Its ground state electronic configuration is [Ar] 3d⁷ 4s².
- For Co^(+3), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d⁶.
- **Conclusion**: 4 unpaired electrons → Colored.

- **(iv) Fe^(+2)**:
- Iron (Fe) has an atomic number of 26. Its ground state electronic configuration is [Ar] 3d⁶ 4s².
- For Fe^(+2), we remove 2 electrons (from 4s), resulting in [Ar] 3d⁶.
- **Conclusion**: 4 unpaired electrons → Colored.

3. **Final Decision on Colorless Ions**:
- From the analysis, the ions that are colorless are:
- (i) Ti^(+4)
- (ii) Cu^(+1)

### Summary of Colorless Ions:
- The colorless ions among the options are:
- (i) Ti^(+4)
- (ii) Cu^(+1)