Paramagnetism is not exhibited by

Options

A. `CuSO_(4)5H_(2)O`
B. `CuCl_(2)5H_(2)O`
C. `CuI`
D. `NiSO_(4).6H_(2)O`

Answer

To determine which compound does not exhibit paramagnetism, we need to analyze the electronic configurations of the given compounds and identify the presence of unpaired electrons. Here’s a step-by-step solution:

### Step 1: Understand Paramagnetism
Paramagnetism occurs in species that have unpaired electrons. If all electrons are paired, the species is diamagnetic and does not exhibit paramagnetism.

### Step 2: Analyze Copper Compounds
We will examine the following compounds:
1. Copper(II) sulfate (CuSO₄·5H₂O)
2. Copper(II) chloride (CuCl₂·2H₂O)
3. Copper(I) iodide (CuI)
4. Nickel(II) sulfate (NiSO₄)

### Step 3: Determine the Oxidation State and Electron Configuration

1. **Copper(II) sulfate (CuSO₄·5H₂O)**:
- Copper is in the +2 oxidation state.
- Electron configuration: Cu²⁺ = [Ar] 3d⁹
- This has 1 unpaired electron (3d⁹).

2. **Copper(II) chloride (CuCl₂·2H₂O)**:
- Copper is also in the +2 oxidation state.
- Electron configuration: Cu²⁺ = [Ar] 3d⁹
- This also has 1 unpaired electron (3d⁹).

3. **Copper(I) iodide (CuI)**:
- Copper is in the +1 oxidation state.
- Electron configuration: Cu⁺ = [Ar] 3d¹⁰
- This has all paired electrons (3d¹⁰), hence it is **not paramagnetic**.

4. **Nickel(II) sulfate (NiSO₄)**:
- Nickel is in the +2 oxidation state.
- Electron configuration: Ni²⁺ = [Ar] 3d⁸
- This has 2 unpaired electrons (3d⁸), hence it is paramagnetic.

### Step 4: Conclusion
From the analysis, we find that **Copper(I) iodide (CuI)** does not exhibit paramagnetism because it has a completely filled d-subshell (3d¹⁰) with no unpaired electrons.

### Final Answer:
**Copper(I) iodide (CuI)** does not exhibit paramagnetism.

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