In any transition series, from left to right, the d-orbitals are progressively filled and their properties vary accordingly.
Q. Which of the following pair of compounds is expected to exhibit same colour in aqueous solution?

Options

A. `FeCl_2,CuCl_2`
B. `VOCl_2,CuCl_2`
C. `VOCl_2,FeCl_2`
D. `FeCl_2,MnCl_2`

Explanation

To determine which pair of compounds is expected to exhibit the same color in aqueous solution, we need to analyze the electronic configurations of the transition metals involved and their oxidation states. The color of transition metal complexes in solution is primarily due to d-d transitions, which depend on the number of unpaired electrons in the d-orbitals.

### Step-by-Step Solution:

1. **Identify the Transition Metals and Their Oxidation States**:
- We have the following transition metals: Iron (Fe), Copper (Cu), Vanadium (V), and Manganese (Mn).
- The oxidation states given in the question are:
- Fe: +2
- Cu: +2
- V: +4
- Mn: +2

2. **Write the Electronic Configurations**:
- For each transition metal, we need to determine the electronic configuration in the given oxidation states:
- **Iron (Fe)**:
- Atomic number = 26
- Electronic configuration: [Ar] 3d6 4s2
- For Fe²⁺: Remove 2 electrons from 4s and 3d orbitals → [Ar] 3d6
- **Copper (Cu)**:
- Atomic number = 29
- Electronic configuration: [Ar] 3d10 4s1
- For Cu²⁺: Remove 2 electrons (1 from 4s and 1 from 3d) → [Ar] 3d9
- **Vanadium (V)**:
- Atomic number = 23
- Electronic configuration: [Ar] 3d3 4s2
- For V⁴⁺: Remove 4 electrons (2 from 4s and 2 from 3d) → [Ar] 3d1
- **Manganese (Mn)**:
- Atomic number = 25
- Electronic configuration: [Ar] 3d5 4s2
- For Mn²⁺: Remove 2 electrons from 4s → [Ar] 3d5

3. **Determine the Number of Unpaired Electrons**:
- The number of unpaired electrons in each oxidation state is crucial for determining the color:
- Fe²⁺ (3d6): 4 unpaired electrons
- Cu²⁺ (3d9): 1 unpaired electron
- V⁴⁺ (3d1): 1 unpaired electron
- Mn²⁺ (3d5): 5 unpaired electrons

4. **Compare the Number of Unpaired Electrons**:
- Now we compare the unpaired electrons for the pairs given in the options:
- **Option 1**: Fe²⁺ (4 unpaired) and Cu²⁺ (1 unpaired) → Different colors
- **Option 2**: V⁴⁺ (1 unpaired) and Cu²⁺ (1 unpaired) → Same color
- **Option 3**: V⁴⁺ (1 unpaired) and Fe²⁺ (4 unpaired) → Different colors
- **Option 4**: Mn²⁺ (5 unpaired) and Fe²⁺ (4 unpaired) → Different colors

5. **Conclusion**:
- The only pair of compounds that exhibit the same color in aqueous solution due to having the same number of unpaired electrons is:
- **Option 2**: V⁴⁺ and Cu²⁺.

### Final Answer:
The pair of compounds expected to exhibit the same color in aqueous solution is **V⁴⁺ and Cu²⁺**.