The colourless species is

Options

A. `VCl_3`
B. `VOSO_4`
C. `Na_3VO_4`
D. `[V(H_2O)_6]SO_(4).H_2O`

Explanation

To determine which of the given species is colorless, we need to analyze the oxidation states of vanadium in each compound and check for unpaired electrons. The presence of unpaired electrons is what typically gives transition metal complexes their color.

### Step-by-Step Solution:

1. **Identify the Compounds**: We have four compounds to analyze:
- VCl3
- VOSO4
- Na3VO4
- [V(SO4)6]•xH2O

2. **Determine the Oxidation State of Vanadium in Each Compound**:
- **VCl3**:
- Let the oxidation state of vanadium be \( x \).
- The equation is \( x + 3(-1) = 0 \) (since Cl is -1).
- Solving gives \( x = +3 \).

- **VOSO4**:
- Let the oxidation state of vanadium be \( x \).
- The equation is \( x - 2 + 6 - 8 = 0 \) (O is -2, S is +6, and there are 4 O).
- Solving gives \( x = +4 \).

- **Na3VO4**:
- Let the oxidation state of vanadium be \( x \).
- The equation is \( 3(+1) + x - 8 = 0 \) (Na is +1, and there are 4 O).
- Solving gives \( x = +5 \).

- **[V(SO4)6]•xH2O**:
- Let the oxidation state of vanadium be \( x \).
- The equation is \( x + 6(-2) = 0 \) (SO4 is -2).
- Solving gives \( x = +2 \).

3. **Check for Unpaired Electrons**:
- **VCl3 (Vanadium +3)**:
- Electron configuration: \( [Ar] 3d^3 \) (3 unpaired electrons).
- **VOSO4 (Vanadium +4)**:
- Electron configuration: \( [Ar] 3d^2 \) (2 unpaired electrons).
- **Na3VO4 (Vanadium +5)**:
- Electron configuration: \( [Ar] 3d^0 \) (0 unpaired electrons).
- **[V(SO4)6]•xH2O (Vanadium +2)**:
- Electron configuration: \( [Ar] 3d^4 \) (4 unpaired electrons).

4. **Conclusion**:
- The only species with no unpaired electrons is **Na3VO4** (Vanadium in +5 oxidation state), making it the colorless species.

### Final Answer:
The colorless species is **Na3VO4**.