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 of the following is paramagnetic as well as coloured ion?

Options

A. `Cu^(o+)`
B. `Cu^(2+)`
C. `Sc^(3+)`
D. `Zn^(2+)`

Answer

To determine which of the given transition metal ions is both paramagnetic and colored, we need to analyze the electronic configurations of the ions and their ability to undergo d-d transitions. Let's break down the solution step by step.

### Step 1: Understand the Concept of Paramagnetism and Color
- **Paramagnetic substances** have unpaired electrons. The presence of unpaired electrons allows for d-d transitions, which are responsible for the color in transition metal complexes.
- **Colored ions** in transition metals arise from d-d transitions where electrons are excited from one d-orbital to another.

### Step 2: Analyze the Given Options
We will analyze the electronic configurations of the transition metal ions provided in the question.

1. **Cu⁺ (Copper(I))**
- Atomic number of Cu = 29
- Electronic configuration: [Ar] 3d¹⁰ 4s²
- In Cu⁺, it loses one electron, resulting in: [Ar] 3d¹⁰
- **Conclusion**: Fully filled d-orbitals (3d¹⁰) mean no unpaired electrons. Therefore, Cu⁺ is colorless and not paramagnetic.

2. **Cu²⁺ (Copper(II))**
- For Cu²⁺, it loses two electrons, resulting in: [Ar] 3d⁹
- This configuration has 9 electrons in the d-orbitals, meaning there is 1 unpaired electron.
- **Conclusion**: Cu²⁺ is paramagnetic (due to the unpaired electron) and can undergo d-d transitions, making it colored.

3. **Sc³⁺ (Scandium(III))**
- Atomic number of Sc = 21
- Electronic configuration: [Ar] 3d¹ 4s²
- In Sc³⁺, it loses three electrons, resulting in: [Ar] 3d⁰
- **Conclusion**: No d-electrons are present, making Sc³⁺ colorless and not paramagnetic.

4. **Zn²⁺ (Zinc(II))**
- Atomic number of Zn = 30
- Electronic configuration: [Ar] 3d¹⁰ 4s²
- In Zn²⁺, it loses two electrons, resulting in: [Ar] 3d¹⁰
- **Conclusion**: Fully filled d-orbitals (3d¹⁰) mean no unpaired electrons. Therefore, Zn²⁺ is colorless and not paramagnetic.

### Step 3: Final Conclusion
After analyzing all the options, the only ion that is both paramagnetic and colored is **Cu²⁺**.

### Summary of the Answer
The transition metal ion that is paramagnetic as well as colored is **Cu²⁺**.