Among the following the compound that is both paramagnetic and coloured is

Options

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

Explanation

To determine which compound among the given options is both paramagnetic and colored, we need to analyze each compound for the presence of unpaired electrons, as paramagnetism is due to unpaired electrons, and compounds with unpaired electrons often exhibit color due to d-d transitions.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- We have four options to consider:
1. K2Cr2O7 (Potassium dichromate)
2. CoSO4 (Cobalt(II) sulfate)
3. Ti(SO4)2 (Titanium(IV) sulfate)
4. CuSO4 (Copper(I) sulfate)

2. **Analyze Option 1: K2Cr2O7**
- **Oxidation State of Chromium:** In K2Cr2O7, the oxidation state of Cr is +6.
- **Electronic Configuration of Cr:** The atomic number of Cr is 24, with the configuration: [Ar] 3d^5 4s^1.
- **After losing 6 electrons:** The configuration becomes [Ar], resulting in 0 unpaired electrons.
- **Conclusion:** K2Cr2O7 is not paramagnetic and therefore not colored.

3. **Analyze Option 2: CoSO4**
- **Oxidation State of Cobalt:** In CoSO4, the oxidation state of Co is +2.
- **Electronic Configuration of Co:** The atomic number of Co is 27, with the configuration: [Ar] 3d^7 4s^2.
- **After losing 2 electrons:** The configuration becomes 3d^7 (with 3 unpaired electrons).
- **Conclusion:** CoSO4 is paramagnetic and likely colored.

4. **Analyze Option 3: Ti(SO4)2**
- **Oxidation State of Titanium:** In Ti(SO4)2, the oxidation state of Ti is +4.
- **Electronic Configuration of Ti:** The atomic number of Ti is 22, with the configuration: [Ar] 3d^2 4s^2.
- **After losing 4 electrons:** The configuration becomes [Ar], resulting in 0 unpaired electrons.
- **Conclusion:** Ti(SO4)2 is not paramagnetic and therefore not colored.

5. **Analyze Option 4: CuSO4**
- **Oxidation State of Copper:** In CuSO4, the oxidation state of Cu is +1.
- **Electronic Configuration of Cu:** The atomic number of Cu is 29, with the configuration: [Ar] 3d^10 4s^1.
- **After losing 1 electron:** The configuration becomes 3d^10 (with 0 unpaired electrons).
- **Conclusion:** CuSO4 is not paramagnetic and therefore not colored.

### Final Conclusion:
The only compound that is both paramagnetic and colored is **CoSO4** (Cobalt(II) sulfate).

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