Compound that is both paramagnetic and coloured is:

Options

A. `K_(2)Cr_(2)O_(7)`
B. `(NH_(4))_(2)[TiCl_(6)]`
C. `VOSO_(4)`
D. `K_(3)[Cu(CN)_(4)]`

Explanation

To determine which compound is both paramagnetic and colored, we will analyze the given compounds one by one, focusing on their oxidation states, electronic configurations, and the presence of unpaired electrons.

### Step-by-Step Solution:

1. **Analyze K2Cr2O7 (Potassium Dichromate)**:
- **Oxidation State**: Chromium is in the +6 oxidation state.
- **Electronic Configuration**: The electronic configuration of Cr in +6 state is Ar 3d^0 4s^0.
- **Unpaired Electrons**: There are no unpaired electrons (3d^0).
- **Magnetic Behavior**: Since there are no unpaired electrons, K2Cr2O7 is **diamagnetic**.
- **Color**: It is colored (orange/red).

**Conclusion**: K2Cr2O7 is colored but **not paramagnetic**.

2. **Analyze NH4 whole twice TiCl6 (Ammonium Titanium(IV) Chloride)**:
- **Oxidation State**: Titanium is in the +4 oxidation state.
- **Electronic Configuration**: The electronic configuration of Ti in +4 state is Ar 3d^0 4s^0.
- **Unpaired Electrons**: There are no unpaired electrons (3d^0).
- **Magnetic Behavior**: Since there are no unpaired electrons, NH4 whole twice TiCl6 is **diamagnetic**.
- **Color**: It is colored (yellow).

**Conclusion**: NH4 whole twice TiCl6 is colored but **not paramagnetic**.

3. **Analyze Vanadyl Sulfate (VO(SO4))**:
- **Oxidation State**: Vanadium is in the +4 oxidation state.
- **Electronic Configuration**: The electronic configuration of V in +4 state is Ar 3d^1 4s^0.
- **Unpaired Electrons**: There is 1 unpaired electron (3d^1).
- **Magnetic Behavior**: Since there is one unpaired electron, Vanadyl sulfate is **paramagnetic**.
- **Color**: It is colored (blue).

**Conclusion**: Vanadyl sulfate is both **paramagnetic and colored**.

4. **Analyze CuSO4 (Copper(I) Sulfate)**:
- **Oxidation State**: Copper is in the +1 oxidation state.
- **Electronic Configuration**: The electronic configuration of Cu in +1 state is Ar 3d^10 4s^0.
- **Unpaired Electrons**: There are no unpaired electrons (3d^10).
- **Magnetic Behavior**: Since there are no unpaired electrons, CuSO4 is **diamagnetic**.
- **Color**: It is colored (white).

**Conclusion**: CuSO4 is colored but **not paramagnetic**.

### Final Answer:
The compound that is both paramagnetic and colored is **Vanadyl Sulfate (VO(SO4))**.