`[Cu(NH_3)_4]^(2+)` ion is coloured while `[Cu(CN)_4]^(3-)` ion is colourless why ?

Explanation

To understand why the `[Cu(NH_3)_4]^{2+}` ion is colored while the `[Cu(CN)_4]^{3-}` ion is colorless, we need to analyze the electronic configurations and the presence of unpaired electrons in these complexes.

### Step-by-Step Solution:

1. **Identify the oxidation states of copper in both complexes**:
- In `[Cu(NH_3)_4]^{2+}`, copper is in the +2 oxidation state.
- In `[Cu(CN)_4]^{3-}`, copper is in the +1 oxidation state.

2. **Determine the electronic configuration of copper in each oxidation state**:
- For Cu in the +2 state (Cu²⁺):
- The atomic number of copper (Cu) is 29. The electron configuration of neutral copper is `[Ar] 4s^1 3d^{10}`.
- When copper loses two electrons to become Cu²⁺, it loses the 4s electron and one 3d electron, resulting in the configuration: `3d^9`.
- For Cu in the +1 state (Cu⁺):
- In the +1 oxidation state, copper loses only one electron, which is the 4s electron, resulting in the configuration: `3d^{10}`.

3. **Analyze the presence of unpaired electrons**:
- In the `[Cu(NH_3)_4]^{2+}` complex (Cu²⁺ with `3d^9`):
- The `3d^9` configuration means there is one unpaired electron. This unpaired electron allows for d-d transitions, which are responsible for the absorption of visible light, thus imparting color to the complex.
- In the `[Cu(CN)_4]^{3-}` complex (Cu⁺ with `3d^{10}`):
- The `3d^{10}` configuration has all electrons paired. Since there are no unpaired electrons, there are no d-d transitions possible. As a result, this complex does not absorb visible light and appears colorless.

4. **Conclusion**:
- The presence of unpaired electrons in the `[Cu(NH_3)_4]^{2+}` complex allows for d-d transitions, resulting in color. In contrast, the `[Cu(CN)_4]^{3-}` complex, with no unpaired electrons, does not undergo d-d transitions and is therefore colorless.