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Paramagnesim is not exhibited by:...

Paramagnesim is not exhibited by:

A

`CuSO_(4)*5H_(2)O`

B

`CuCl_(2)*5H_(2)O`

C

`CuI`

D

`NiSO_(4)*4H_(2)O`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which compound does not exhibit paramagnetism, we need to analyze the oxidation states and electronic configurations of the given compounds. Paramagnetism arises from the presence of unpaired electrons in the electronic configuration of an atom or ion. ### Step-by-Step Solution: 1. **Understanding Paramagnetism**: - Paramagnetism occurs due to the presence of unpaired electrons in an atom or ion. If all electrons are paired, the substance is diamagnetic and does not exhibit paramagnetism. 2. **Analyzing Each Compound**: - We will analyze the given compounds one by one to determine their oxidation states and electronic configurations. 3. **Option 1: CuSO4·5H2O**: - **Oxidation State**: Copper (Cu) has a +2 oxidation state because sulfate (SO4) has a -2 charge and water is neutral. - **Electronic Configuration**: For Cu²⁺, the electronic configuration is Argon (Ar) 3d⁹. - **Unpaired Electrons**: There is 1 unpaired electron in 3d⁹, so this compound is **paramagnetic**. 4. **Option 2: CuCl2·5H2O**: - **Oxidation State**: Copper (Cu) also has a +2 oxidation state in this compound. - **Electronic Configuration**: For Cu²⁺, it is again Argon (Ar) 3d⁹. - **Unpaired Electrons**: There is 1 unpaired electron in 3d⁹, so this compound is **paramagnetic**. 5. **Option 3: CuI**: - **Oxidation State**: In copper iodide, copper (Cu) has a +1 oxidation state. - **Electronic Configuration**: For Cu⁺, the electronic configuration is Argon (Ar) 3d¹⁰. - **Unpaired Electrons**: All electrons are paired in 3d¹⁰, so this compound is **not paramagnetic** (it is diamagnetic). 6. **Option 4: NiSO4·4H2O**: - **Oxidation State**: Nickel (Ni) has a +2 oxidation state. - **Electronic Configuration**: For Ni²⁺, the electronic configuration is Argon (Ar) 3d⁸. - **Unpaired Electrons**: There are 2 unpaired electrons in 3d⁸, so this compound is **paramagnetic**. ### Conclusion: The compound that does not exhibit paramagnetism is **CuI (Copper Iodide)**.
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