The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range.
Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour.
The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`.
Q. Which is a coloured ion?

Options

A. `[Cr(H_2O)_6]^(3+)`
B. `[Cu(CN)_4]^(3-)`
C. `[Ti(H_2O)_6]^(4+)`
D. `[Sc(H_2O)_6]^(3+)`

Answer

To determine which ion is colored among the given options, we need to analyze the electronic configurations and the presence of unpaired electrons in each transition metal ion. The presence of unpaired electrons allows for d-d transitions, which are responsible for the color of the ion.

### Step-by-Step Solution:

1. **Identify the ions and their oxidation states:**
- The ions given are:
1. Chromium in \( [Cr(H_2O)_6]^{3+} \)
2. Copper in \( [Cu(CN)_4]^{3-} \)
3. Titanium in \( [Ti(H_2O)_6]^{4+} \)
4. Scandium in \( [Sc(H_2O)_6]^{3+} \)

2. **Determine the electronic configuration of each ion:**
- **Chromium \( [Cr(H_2O)_6]^{3+} \)**:
- Chromium (Cr) has an atomic number of 24. Its electronic configuration is \( [Ar] 3d^5 4s^1 \).
- In the \( +3 \) oxidation state, it loses 3 electrons: \( [Ar] 3d^3 \).
- There are 3 unpaired electrons in the \( 3d \) subshell.

- **Copper \( [Cu(CN)_4]^{3-} \)**:
- Copper (Cu) has an atomic number of 29. Its electronic configuration is \( [Ar] 3d^{10} 4s^1 \).
- In the \( +1 \) oxidation state (since \( CN^- \) is a strong field ligand and leads to a low oxidation state), it retains the \( 3d^{10} \) configuration.
- All \( d \) electrons are paired, making it colorless.

- **Titanium \( [Ti(H_2O)_6]^{4+} \)**:
- Titanium (Ti) has an atomic number of 22. Its electronic configuration is \( [Ar] 3d^2 4s^2 \).
- In the \( +4 \) oxidation state, it loses 4 electrons: \( [Ar] 3d^0 \).
- There are no \( d \) electrons present, making it colorless.

- **Scandium \( [Sc(H_2O)_6]^{3+} \)**:
- Scandium (Sc) has an atomic number of 21. Its electronic configuration is \( [Ar] 3d^1 4s^2 \).
- In the \( +3 \) oxidation state, it loses 3 electrons: \( [Ar] 3d^0 \).
- There are no \( d \) electrons present, making it colorless.

3. **Conclusion:**
- The only ion with unpaired electrons is \( [Cr(H_2O)_6]^{3+} \), which has 3 unpaired electrons and can undergo d-d transitions, making it colored.
- Therefore, the colored ion is **Chromium in \( [Cr(H_2O)_6]^{3+} \)**.

### Final Answer:
The colored ion is **Chromium in \( [Cr(H_2O)_6]^{3+} \)**.