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When small amount of SnCl(2) is added to...

When small amount of `SnCl_(2)` is added to a solution of `Hg^(2+)` ions, a silky white precipitate is obtained.The silky white precipitate is due to the formation of:

A

`Hg_(2)Cl_(2)`

B

`SnCl_(4)`

C

`Sn`

D

`Hg`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the formation of a silky white precipitate when `SnCl2` is added to a solution of `Hg^(2+)` ions, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: The reactants in this scenario are `SnCl2` (tin(II) chloride) and `Hg^(2+)` ions (mercury(II) ions). 2. **Understand the Reaction**: When `SnCl2` is added to a solution containing `Hg^(2+)` ions, a reaction occurs. The tin(II) ions can reduce the mercury(II) ions. 3. **Determine the Possible Products**: The possible products from the reaction are: - `Hg2Cl2` (mercury(I) chloride) - `SnCl4` (tin(IV) chloride) - `Sn` (tin) - `Hg` (mercury) 4. **Focus on the Precipitate**: The question specifically mentions a "silky white precipitate." We need to identify which of the products has these characteristics. 5. **Identify the Silky White Precipitate**: Among the possible products, `Hg2Cl2` is known to form a white precipitate and has a silky appearance. This is a key characteristic of mercury(I) chloride. 6. **Conclusion**: Therefore, when small amounts of `SnCl2` are added to a solution of `Hg^(2+)` ions, the silky white precipitate formed is due to the formation of `Hg2Cl2`. ### Final Answer: The silky white precipitate is due to the formation of `Hg2Cl2`. ---
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