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In qualitative analysis of group I radic...

In qualitative analysis of group I radicals, a white precipitate is formed which is insoluble in boiling water but when treated with `NH_4`OH it turns black. The precipitate may be

A

`PbCl_2`

B

AgCI

C

`HgCl_2`

D

`Hg_2Cl_2`

Text Solution

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The correct Answer is:
To solve the question regarding the white precipitate formed in qualitative analysis of group I radicals, we will analyze each of the options provided step by step. ### Step-by-Step Solution: 1. **Understanding the Problem:** - We are looking for a white precipitate that is insoluble in boiling water. - Upon treatment with `NH4OH`, the precipitate turns black. 2. **Analyzing the Options:** - Let's denote the options as follows: - A: PbCl2 (Lead(II) chloride) - B: AgCl (Silver chloride) - C: HgCl2 (Mercury(II) chloride) - D: Hg2Cl2 (Mercurous chloride) 3. **Option A: PbCl2** - **Solubility Check:** PbCl2 is soluble in boiling water. - **Conclusion:** This cannot be our answer since we need a precipitate that is insoluble in boiling water. 4. **Option B: AgCl** - **Solubility Check:** AgCl is insoluble in boiling water. - **Reaction with NH4OH:** - When AgCl is treated with NH4OH, it forms a soluble complex, [Ag(NH3)2]+ and Cl-. - **Conclusion:** This cannot be our answer since it does not produce a black precipitate. 5. **Option C: HgCl2** - **Solubility Check:** HgCl2 is also soluble in boiling water. - **Conclusion:** This cannot be our answer since we need a precipitate that is insoluble in boiling water. 6. **Option D: Hg2Cl2** - **Solubility Check:** Hg2Cl2 is insoluble in boiling water. - **Reaction with NH4OH:** - When Hg2Cl2 is treated with NH4OH, it forms Hg(NH2)Cl and elemental mercury (Hg), which is black. - **Conclusion:** This fits the criteria of forming a black precipitate upon treatment with NH4OH. 7. **Final Answer:** - The white precipitate that is insoluble in boiling water and turns black when treated with NH4OH is **Hg2Cl2 (Mercurous chloride)**.
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