Which pair of compounds is expected to show similar colour in aqueous medium?

Options

A. `FeCl_(2) and CuCl_(2)`
B. `VOCl_(2) and CuCl_(2)`
C. `VOCl_(2) and FeCl_(2)`
D. `FeCl_(2) and MnCl_(2)`

Answer

To determine which pair of compounds is expected to show similar color in aqueous medium, we need to analyze the electronic configurations of the transition metal ions present in the compounds. The color of transition metal compounds in solution is primarily due to the d-d transitions of unpaired electrons in the d orbitals.

### Step-by-Step Solution:

1. **Identify the Compounds**: We need to analyze the compounds given in the options. For this solution, let's consider VOCl2 (Vanadium(IV) chloride) and CuCl (Copper(I) chloride).

2. **Determine the Oxidation States**:
- In VOCl2, vanadium is in the +4 oxidation state (V^4+).
- In CuCl, copper is in the +1 oxidation state (Cu^+).

3. **Find the Electronic Configuration**:
- For Vanadium (atomic number 23):
- Ground state configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²
- For V^4+: Remove 4 electrons (2 from 4s and 2 from 3d): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ (1 unpaired electron).
- For Copper (atomic number 29):
- Ground state configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
- For Cu^+: Remove 1 electron from 4s: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ (0 unpaired electrons).

4. **Count the Number of Unpaired Electrons**:
- V^4+ has 1 unpaired electron.
- Cu^+ has 0 unpaired electrons.

5. **Conclusion on Color Similarity**:
- Since the number of unpaired electrons in V^4+ is 1 and in Cu^+ is 0, they will not show similar colors.
- However, if we consider Cu^2+ instead (which has 1 unpaired electron), it would then match with V^4+.

6. **Final Answer**: The pair of compounds that is expected to show similar color in aqueous medium is VOCl2 (V^4+) and CuCl2 (Cu^2+).