For `Zn^(2), Ni^(2+), Cu^(2+) and Cr^(2+)` which of the following statements is correct ?

Options

A. 1. Only `Zn^(2+)` is colourless and `Ni^(2+) and Cr^(2+)` are coloured.
B. 2. All the ions are coloured.
C. 3. All the ions are colourless.
D. 4. `Zn^(2+) and Cu^(2+)` are colourless while `Ni^(2+) and Cr^(2+)` are coloured

Explanation

To determine which of the statements regarding the ions Zn²⁺, Ni²⁺, Cu²⁺, and Cr²⁺ is correct, we need to analyze the electronic configurations of these ions and identify whether they have unpaired electrons in their d-orbitals, as this will indicate whether they are colored or colorless.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers:**
- Zinc (Zn): Atomic number = 30
- Nickel (Ni): Atomic number = 28
- Copper (Cu): Atomic number = 29
- Chromium (Cr): Atomic number = 24

2. **Write the Electronic Configurations:**
- For Zn (Z=30):
- Ground state: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰
- For Zn²⁺: Remove 2 electrons (from 4s²):
- Zn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ (3d is fully filled)
- For Ni (Z=28):
- Ground state: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸
- For Ni²⁺: Remove 2 electrons (from 4s²):
- Ni²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ (3d has unpaired electrons)
- For Cu (Z=29):
- Ground state: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰
- For Cu²⁺: Remove 2 electrons (1 from 4s² and 1 from 3d¹⁰):
- Cu²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ (3d has unpaired electrons)
- For Cr (Z=24):
- Ground state: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
- For Cr²⁺: Remove 2 electrons (from 4s²):
- Cr²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ (3d has unpaired electrons)

3. **Determine Color or Colorlessness:**
- **Zn²⁺:** 3d¹⁰ (fully filled d-orbitals, colorless)
- **Ni²⁺:** 3d⁸ (unpaired electrons, colored)
- **Cu²⁺:** 3d⁹ (unpaired electrons, colored)
- **Cr²⁺:** 3d⁴ (unpaired electrons, colored)

4. **Conclusion:**
- The only ion that is colorless is Zn²⁺, while Ni²⁺, Cu²⁺, and Cr²⁺ are colored due to the presence of unpaired electrons.

### Final Answer:
The correct statement is: **Only Zn²⁺ is colorless, and Ni²⁺, Cu²⁺, and Cr²⁺ are colored.**