Among the following, the species that is both paramagnetic and coloured is :

Options

A. `[MnO_(4)]^(2-)`
B. `K_(4)[V(CN)_(6)]`
C. `[VO_(4)]^(3-)`
D. `CrO_(2)Cl_(2)`

Answer

To solve the problem of identifying the species that is both paramagnetic and colored, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Identify the Species**: The options given are:
- A) MnO4²⁻
- B) K4VCN6
- C) VO4³⁻
- D) CrO2Cl2

2. **Analyze Option A: MnO4²⁻**
- **Determine Oxidation State**: Let the oxidation state of manganese be \( x \).
\[ x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6 \]
- **Electronic Configuration**: Manganese (Mn) has an atomic number of 25, so its electronic configuration is:
\[ \text{Mn: } [Ar] 3d^5 4s^2 \]
- In the +6 oxidation state, Mn loses 6 electrons:
\[ \text{Mn}^{+6}: [Ar] 3d^1 \]
- **Paramagnetism**: There is 1 unpaired electron in the 3d orbital, so MnO4²⁻ is paramagnetic.
- **Color**: MnO4²⁻ exhibits color due to metal-to-ligand charge transfer.
- **Conclusion**: MnO4²⁻ is both paramagnetic and colored.

3. **Analyze Option B: K4VCN6**
- **Determine Oxidation State**: Let the oxidation state of vanadium be \( x \).
\[ 4(+1) + x + 6(-1) = 0 \implies 4 + x - 6 = 0 \implies x = +2 \]
- **Electronic Configuration**: Vanadium (V) has an atomic number of 23:
\[ \text{V: } [Ar] 3d^3 4s^2 \]
- In the +2 oxidation state:
\[ \text{V}^{+2}: [Ar] 3d^3 \]
- **Paramagnetism**: There are 3 unpaired electrons, so K4VCN6 is paramagnetic.
- **Color**: However, the complex is not colored due to lack of allowed electronic transitions.
- **Conclusion**: K4VCN6 is paramagnetic but not colored.

4. **Analyze Option C: VO4³⁻**
- **Determine Oxidation State**: Let the oxidation state of vanadium be \( x \).
\[ x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5 \]
- **Electronic Configuration**: In the +5 oxidation state:
\[ \text{V}^{+5}: [Ar] 3d^0 \]
- **Paramagnetism**: There are 0 unpaired electrons, so VO4³⁻ is diamagnetic.
- **Conclusion**: VO4³⁻ is neither paramagnetic nor colored.

5. **Analyze Option D: CrO2Cl2**
- **Determine Oxidation State**: Let the oxidation state of chromium be \( x \).
\[ x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6 \]
- **Electronic Configuration**: In the +6 oxidation state:
\[ \text{Cr}^{+6}: [Ar] 3d^0 4s^0 \]
- **Paramagnetism**: There are 0 unpaired electrons, so CrO2Cl2 is diamagnetic.
- **Conclusion**: CrO2Cl2 is neither paramagnetic nor colored.

### Final Conclusion:
The only species that is both paramagnetic and colored is **MnO4²⁻**.