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 determine which species among the given options is both paramagnetic and colored, we need to analyze the oxidation states and electronic configurations of the compounds.

### Step-by-Step Solution:

1. **Identify the Compounds:**
We have the following species to analyze:
- MnO4^-
- VCl4
- VO4^3-
- CrO2Cl2

2. **Calculate the Oxidation States:**
- **For MnO4^-:**
Let the oxidation state of Mn be \( x \).
\[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \]
- **For VCl4:**
Let the oxidation state of V be \( x \).
\[ x + 4(-1) = 0 \implies x - 4 = 0 \implies x = +4 \]
- **For VO4^3-:**
Let the oxidation state of V be \( x \).
\[ x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5 \]
- **For CrO2Cl2:**
Let the oxidation state of Cr be \( x \).
\[ x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6 \]

3. **Write the Electronic Configurations:**
- **For Mn (Z=25) in +7 state:**
Ground state: [Ar] 3d^5 4s^2
\[ \text{Mn}^{7+}: [Ar] \quad (3d^0) \]
(No unpaired electrons)

- **For V (Z=23) in +4 state:**
Ground state: [Ar] 4s^2 3d^3
\[ \text{V}^{4+}: [Ar] \quad (3d^2) \]
(2 unpaired electrons)

- **For V (Z=23) in +5 state:**
Ground state: [Ar] 4s^2 3d^3
\[ \text{V}^{5+}: [Ar] \quad (3d^1) \]
(1 unpaired electron)

- **For Cr (Z=24) in +6 state:**
Ground state: [Ar] 4s^2 3d^4
\[ \text{Cr}^{6+}: [Ar] \quad (3d^0) \]
(No unpaired electrons)

4. **Determine Paramagnetism and Color:**
- **MnO4^-**: No unpaired electrons → Not paramagnetic, not colored.
- **VCl4**: 2 unpaired electrons → Paramagnetic and colored.
- **VO4^3-**: 1 unpaired electron → Paramagnetic and colored.
- **CrO2Cl2**: No unpaired electrons → Not paramagnetic, not colored.

### Conclusion:
The species that is both paramagnetic and colored are **VCl4** and **VO4^3-**.