Out of the following ions `Ti^(3+), V^(3+), Cu^(+), Sc^(3+), Mn^(3+) and Co^(2+)` the colourless ions will be

Options

A. `Cu^(+), Sc^(3+)`
B. `Ti^(3+), V^(3+)`
C. `Cu^(+), Co^(2+)`
D. `Sc^(3+), Fe^(3+)`

Answer

To determine which of the given ions are colorless, we need to analyze the electronic configurations of each ion and check for the presence of unpaired electrons. Colorless ions are typically those that are diamagnetic, meaning they have all paired electrons.

### Step-by-Step Solution:

1. **Identify the Ions:**
The ions we need to analyze are:
- Ti³⁺
- V³⁺
- Cu⁺
- Sc³⁺
- Mn³⁺
- Co²⁺

2. **Determine the Electronic Configuration of Each Ion:**
- **Ti³⁺ (Titanium):**
- Atomic number = 22
- Electronic configuration of Ti: [Ar] 4s² 3d²
- For Ti³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d¹ (1 unpaired electron)
- **Conclusion:** Paramagnetic (colored)

- **V³⁺ (Vanadium):**
- Atomic number = 23
- Electronic configuration of V: [Ar] 4s² 3d³
- For V³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d² (2 unpaired electrons)
- **Conclusion:** Paramagnetic (colored)

- **Cu⁺ (Copper):**
- Atomic number = 29
- Electronic configuration of Cu: [Ar] 4s¹ 3d¹⁰
- For Cu⁺, we remove 1 electron (from 4s):
- Configuration: [Ar] 3d¹⁰ (all paired)
- **Conclusion:** Diamagnetic (colorless)

- **Sc³⁺ (Scandium):**
- Atomic number = 21
- Electronic configuration of Sc: [Ar] 4s² 3d¹
- For Sc³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d⁰ (all paired)
- **Conclusion:** Diamagnetic (colorless)

- **Mn³⁺ (Manganese):**
- Atomic number = 25
- Electronic configuration of Mn: [Ar] 4s² 3d⁵
- For Mn³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d⁴ (4 unpaired electrons)
- **Conclusion:** Paramagnetic (colored)

- **Co²⁺ (Cobalt):**
- Atomic number = 27
- Electronic configuration of Co: [Ar] 4s² 3d⁷
- For Co²⁺, we remove 2 electrons (from 4s):
- Configuration: [Ar] 3d⁷ (3 unpaired electrons)
- **Conclusion:** Paramagnetic (colored)

3. **Identify Colorless Ions:**
From the analysis, the colorless ions are:
- Cu⁺ (colorless)
- Sc³⁺ (colorless)

### Final Answer:
The colorless ions from the given list are **Cu⁺ and Sc³⁺**.

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