Paramagnetism is not exhibited by

Options

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

Answer

To determine which compound does not exhibit paramagnetism, we need to analyze the electronic configurations of the compounds given. Paramagnetism occurs in species that have unpaired electrons. Let's go through the options step by step.

### Step 1: Analyze the Electronic Configuration of Copper
Copper (Cu) has the electronic configuration of:
- Ground state: \( [Ar] 3d^{10} 4s^1 \)
- In the +2 oxidation state (Cu²⁺): \( [Ar] 3d^9 \)

This means that in the +2 state, copper will have one unpaired electron in the 3d subshell.

### Step 2: Analyze Each Compound

1. **CuSO₄·5H₂O (Copper(II) sulfate pentahydrate)**
- In this compound, copper is in the +2 oxidation state (Cu²⁺).
- Electronic configuration: \( [Ar] 3d^9 \)
- Since there is one unpaired electron, this compound **exhibits paramagnetism**.

2. **CuCl₂ (Copper(II) chloride)**
- Similar to CuSO₄·5H₂O, copper is also in the +2 oxidation state (Cu²⁺).
- Electronic configuration: \( [Ar] 3d^9 \)
- This compound also has one unpaired electron, thus it **exhibits paramagnetism**.

3. **CuI (Copper(I) iodide)**
- In this compound, copper is in the +1 oxidation state (Cu⁺).
- Electronic configuration: \( [Ar] 3d^{10} \) (after losing one 4s electron).
- Since all electrons are paired in the 3d subshell, this compound **does not exhibit paramagnetism**.

4. **NiSO₄·4H₂O (Nickel(II) sulfate tetrahydrate)**
- In this compound, nickel is in the +2 oxidation state (Ni²⁺).
- Electronic configuration: \( [Ar] 3d^8 \)
- There are two unpaired electrons, so this compound **exhibits paramagnetism**.

### Conclusion
From the analysis, the compound that does not exhibit paramagnetism is **CuI (Copper(I) iodide)**.