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A chloride salt on addition of alkali so...

A chloride salt on addition of alkali solution gives gas B whicch gives brownn ppt. with nessler's regent. What is A, B and C?

A

`NH_(4)Cl,NH_(3) and HgO*Hg(NH_(2))(NO_(3))`

B

`NH_(4)Cl,NH_(3) and Hg(NH_(3))Cl`

C

`NH_(4)Cl, NH_(3) and HgO*Hg(NH_(2))Cl`

D

`NH_(4)Cl,NH_(3) and HgO*Hg(NH_(2))I`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to identify the chloride salt (A), the gas produced (B), and the brown precipitate formed with Nessler's reagent (C). ### Step-by-Step Solution: 1. **Identify the Chloride Salt (A)**: - The problem states that a chloride salt reacts with an alkali solution. A common chloride salt that fits this description is **Ammonium Chloride (NH4Cl)**. 2. **Reaction with Alkali**: - When Ammonium Chloride (NH4Cl) is treated with an alkali, such as Sodium Hydroxide (NaOH), the following reaction occurs: \[ NH4Cl + NaOH \rightarrow NaCl + NH3 + H2O \] - In this reaction, Ammonium Chloride reacts with Sodium Hydroxide to produce Sodium Chloride (NaCl), Ammonia (NH3), and water (H2O). 3. **Identify the Gas (B)**: - The gas produced in this reaction is **Ammonia (NH3)**. This is the gas (B) mentioned in the question. 4. **Nessler's Reagent Reaction**: - Nessler's reagent, which is a solution of potassium tetraiodomercurate(II) (K2HgI4) in the presence of KOH, reacts with Ammonia (NH3) to form a brown precipitate. The reaction can be represented as: \[ NH3 + K2HgI4 \rightarrow Hg2I2 \cdot NH2 + KOH \] - The brown precipitate formed is **Mercuric Ammonium Iodide (Hg2I2·NH2)**, which is the brown precipitate (C) referred to in the question. 5. **Final Identification**: - Thus, we can summarize: - A (Chloride Salt) = **Ammonium Chloride (NH4Cl)** - B (Gas) = **Ammonia (NH3)** - C (Brown Precipitate) = **Mercuric Ammonium Iodide (Hg2I2·NH2)** ### Conclusion: - The answer to the question is: - A = Ammonium Chloride (NH4Cl) - B = Ammonia (NH3) - C = Mercuric Ammonium Iodide (Hg2I2·NH2)
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