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Kl gives a precipitate with all the cati...

Kl gives a precipitate with all the cations given below. Choose the cation

A

`Ag^+.Hg_(2)^(2+) Pb^(2+)`

B

`Cu^(2+) Zn^(2+) Ni^(2+)`

C

`Na^(+)Ca^(2+) Mg^(2+)`

D

`Ag^+ Ca^(2+) Sc^(2+)`

Text Solution

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The correct Answer is:
To solve the question regarding which cation gives a precipitate with potassium iodide (KI), we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Dissociation of KI**: Potassium iodide (KI) dissociates in solution into potassium ions (K⁺) and iodide ions (I⁻). **Hint**: Remember that KI breaks down into K⁺ and I⁻ in aqueous solution. 2. **Identify the Reaction with Cations**: The iodide ions (I⁻) will react with various cations to form precipitates. We need to check which cations form insoluble iodides. 3. **Check Each Option**: - **Option A**: Contains silver (Ag⁺), lead (Pb²⁺), and other cations. - Ag⁺ + I⁻ → AgI (yellow precipitate) - Pb²⁺ + I⁻ → PbI₂ (yellow precipitate) - Other cations in this option also form precipitates. - **Option B**: Contains copper (Cu²⁺), zinc (Zn²⁺), and nickel (Ni²⁺). - Cu²⁺ + I⁻ → CuI (white precipitate) but Zn²⁺ and Ni²⁺ do not form precipitates with I⁻. - **Option C**: Contains sodium (Na⁺), calcium (Ca²⁺), and magnesium (Mg²⁺). - None of these cations form precipitates with I⁻. - **Option D**: Contains silver (Ag⁺), calcium (Ca²⁺), and scandium (Sc³⁺). - Ag⁺ + I⁻ → AgI (yellow precipitate), but Ca²⁺ and Sc³⁺ do not form precipitates with I⁻. 4. **Conclusion**: After analyzing all options, **Option A** is the only one where all cations react with iodide ions to form precipitates. **Final Answer**: The correct option is **A**.
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