Which set of ions exhibit specific colours? (Atomic number of Sc = 21, Ti = 22, V=23, Mn = 25, Fe = 26, Ni = 28 Cu = 29 and Zn =30)

Options

A. `Sc^(3+), Ti^(4+), Mn^(3+)`
B. `Sc^(3+), Zn^(2+), Ni^(2+)`
C. `V^(3+), V^(2+), Fe^(3+)`
D. `Ti^(3+), Ti^(4+), Ni^(2+)`

Answer

To determine which set of ions exhibit specific colors, we need to analyze the electron configurations of the transition metal ions provided in the question. The colors exhibited by these ions are primarily due to the presence of unpaired d-electrons, which allow for electronic transitions when they absorb visible light.

### Step-by-Step Solution:

1. **Identify the Electron Configurations**:
We need to write the electron configurations for the ions in question. The relevant transition metals and their oxidation states are as follows:
- Sc³⁺: Argon 3D⁰ (no unpaired electrons)
- Ti⁴⁺: Argon 3D⁰ (no unpaired electrons)
- V³⁺: Argon 3D² (2 unpaired electrons)
- Mn³⁺: Argon 3D⁴ (4 unpaired electrons)
- Fe³⁺: Argon 3D⁵ (5 unpaired electrons)
- Ni²⁺: Argon 3D⁸ (2 unpaired electrons)
- Cu²⁺: Argon 3D⁹ (1 unpaired electron)
- Zn²⁺: Argon 3D¹⁰ (no unpaired electrons)

2. **Determine the Presence of Unpaired Electrons**:
- Ions with unpaired d-electrons can absorb visible light and exhibit color.
- Sc³⁺, Ti⁴⁺, and Zn²⁺ have no unpaired electrons and therefore do not exhibit color.
- V³⁺, Mn³⁺, Fe³⁺, Ni²⁺, and Cu²⁺ have unpaired electrons and can exhibit specific colors.

3. **Conclusion**:
The ions that can exhibit specific colors due to the presence of unpaired electrons are:
- V³⁺
- Mn³⁺
- Fe³⁺
- Ni²⁺
- Cu²⁺

### Final Answer:
The set of ions that exhibit specific colors are V³⁺, Mn³⁺, Fe³⁺, Ni²⁺, and Cu²⁺.

---