The highest degree of paramagnetism is shown by 

Options

A. `CoCl_(2).H_(2)O`
B. `MnSO_(4).4H_(2)O`
C. `FeCl_(2).4H_(2)O`
D. `NiCl_(2).6H_(2)O`

Answer

To determine which compound shows the highest degree of paramagnetism, we need to analyze the number of unpaired electrons in the transition metal ions present in each compound. Paramagnetism is directly related to the presence of unpaired electrons; the more unpaired electrons, the higher the paramagnetism.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Transition Metals**:
- COCl2.H2O (Cobalt)
- MnSO4.4H2O (Manganese)
- FeCl2.4H2O (Iron)
- NiCl2.6H2O (Nickel)

2. **Determine the Oxidation States**:
- For COCl2.H2O: Cobalt (Co) is in +2 oxidation state.
- For MnSO4.4H2O: Manganese (Mn) is in +2 oxidation state.
- For FeCl2.4H2O: Iron (Fe) is in +2 oxidation state.
- For NiCl2.6H2O: Nickel (Ni) is in +2 oxidation state.

3. **Write the Electron Configurations**:
- Cobalt (Co, atomic number 27):
- Neutral: [Ar] 4s² 3d⁷
- Co²⁺: [Ar] 3d⁷ (loses 2 electrons from 4s)
- Manganese (Mn, atomic number 25):
- Neutral: [Ar] 4s² 3d⁵
- Mn²⁺: [Ar] 3d⁵ (loses 2 electrons from 4s)
- Iron (Fe, atomic number 26):
- Neutral: [Ar] 4s² 3d⁶
- Fe²⁺: [Ar] 3d⁶ (loses 2 electrons from 4s)
- Nickel (Ni, atomic number 28):
- Neutral: [Ar] 4s² 3d⁸
- Ni²⁺: [Ar] 3d⁸ (loses 2 electrons from 4s)

4. **Count the Unpaired Electrons**:
- Co²⁺ (3d⁷): 3 unpaired electrons
- Mn²⁺ (3d⁵): 5 unpaired electrons
- Fe²⁺ (3d⁶): 4 unpaired electrons
- Ni²⁺ (3d⁸): 2 unpaired electrons

5. **Compare the Number of Unpaired Electrons**:
- Co²⁺: 3 unpaired electrons
- Mn²⁺: 5 unpaired electrons
- Fe²⁺: 4 unpaired electrons
- Ni²⁺: 2 unpaired electrons

6. **Conclusion**:
- The compound with the highest degree of paramagnetism is **MnSO4.4H2O**, which has **5 unpaired electrons**.

### Final Answer:
The highest degree of paramagnetism is shown by **MnSO4.4H2O**.