Which of the following is a colourless complex ion?

Options

A. `[Cu(NH_(3))_(4)]^(1+)`
B. `[Cr(H_(2)O)_(6)]^(3+)`
C. `[FeF_(6)]^(3-)`
D. `[Mn(H_(2)O)_(6)]^(3+)`

Explanation

To determine which of the following complex ions is colorless, we need to analyze the oxidation states and electronic configurations of the metal ions in each complex. A complex is colorless if it has no unpaired electrons in its d-orbitals.

### Step-by-Step Solution:

1. **Identify the Complex Ions**: The options provided are:
- A) [Cu(NH3)4]⁺
- B) [Cr(H2O)6]³⁺
- C) [FeF6]³⁻
- D) [Mn(H2O)6]³⁺

2. **Analyze Option A: [Cu(NH3)4]⁺**
- Oxidation State: Cu is in +1 oxidation state.
- Electronic Configuration: Cu⁺ has the configuration [Ar] 4s⁰ 3d¹⁰.
- Unpaired Electrons: There are 0 unpaired electrons.
- Conclusion: This complex is colorless.

3. **Analyze Option B: [Cr(H2O)6]³⁺**
- Oxidation State: Cr is in +3 oxidation state.
- Electronic Configuration: Cr³⁺ has the configuration [Ar] 4s⁰ 3d³.
- Unpaired Electrons: There are 3 unpaired electrons.
- Conclusion: This complex is colored.

4. **Analyze Option C: [FeF6]³⁻**
- Oxidation State: Fe is in +3 oxidation state.
- Electronic Configuration: Fe³⁺ has the configuration [Ar] 4s⁰ 3d⁵.
- Unpaired Electrons: There are 5 unpaired electrons.
- Conclusion: This complex is colored.

5. **Analyze Option D: [Mn(H2O)6]³⁺**
- Oxidation State: Mn is in +3 oxidation state.
- Electronic Configuration: Mn³⁺ has the configuration [Ar] 4s⁰ 3d⁴.
- Unpaired Electrons: There are 4 unpaired electrons.
- Conclusion: This complex is colored.

6. **Final Conclusion**: The only complex that is colorless is option A: [Cu(NH3)4]⁺, as it has no unpaired electrons.

### Answer:
The colorless complex ion is **[Cu(NH3)4]⁺** (Option A).

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