Paramagnetism is a property due to the presence of unpaired electrons. In case of transition metals, as they contain unpaired electrons in the `(n-1)` d orbitals,most of the transition metal ions and their compounds are paramagnetic .Para magnetism increases with increases in number of unpaired electrons. Magnetic moment is calculated from spin only formula' `Vz`
`mu=sqrt(n(n+2)) B.M
n="number of unpaired electrons"`
Similarly the colour of the compounds of transition metals may be attributed to the presence of incomplete `(n-1)` d sub-shell. When an electron from a lower energy of d-orbitals is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to complementary colour of the light observed. The frequency of the light absorbed is determined by the nature of the ligand.
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)`

Answer

To determine which pair of compounds is expected to exhibit the same color in aqueous solution, we need to analyze the number of unpaired electrons in each compound. The presence of unpaired electrons in the d-orbitals of transition metals is responsible for their paramagnetic properties and the colors they exhibit.

### Step-by-Step Solution:

1. **Understanding the Question**: The question asks which pair of transition metal compounds exhibits the same color in aqueous solution. The color is influenced by the number of unpaired electrons and the nature of the ligands.

2. **Identifying the Compounds**: The pairs of compounds provided are:
- (i) FeCl₂ and CuCl₂
- (ii) VOCl₂ and CuCl₂
- (iii) VOCl₂ and FeCl₂
- (iv) FeCl₂ and MnCl₂

3. **Finding the Electronic Configurations**:
- **FeCl₂**: Iron (Fe) in +2 oxidation state has the electronic configuration of Ar 3d⁶. This means it has 4 unpaired electrons.
- **CuCl₂**: Copper (Cu) in +2 oxidation state has the electronic configuration of Ar 3d⁹. This means it has 1 unpaired electron.
- **VOCl₂**: Vanadium (V) in +4 oxidation state has the electronic configuration of Ar 3d¹. This means it has 1 unpaired electron.
- **MnCl₂**: Manganese (Mn) in +2 oxidation state has the electronic configuration of Ar 3d⁵. This means it has 5 unpaired electrons.

4. **Counting Unpaired Electrons**:
- **FeCl₂**: 4 unpaired electrons
- **CuCl₂**: 1 unpaired electron
- **VOCl₂**: 1 unpaired electron
- **MnCl₂**: 5 unpaired electrons

5. **Comparing Unpaired Electrons**:
- For the pair **FeCl₂ and CuCl₂**: Different number of unpaired electrons (4 vs. 1).
- For the pair **VOCl₂ and CuCl₂**: Same number of unpaired electrons (1 vs. 1).
- For the pair **VOCl₂ and FeCl₂**: Different number of unpaired electrons (1 vs. 4).
- For the pair **FeCl₂ and MnCl₂**: Different number of unpaired electrons (4 vs. 5).

6. **Conclusion**: The only pair of compounds that exhibits the same color in aqueous solution is **VOCl₂ and CuCl₂**, as they both have 1 unpaired electron.

### Final Answer:
The pair of compounds expected to exhibit the same color in aqueous solution is **VOCl₂ and CuCl₂**.

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