On passing a gas, ‘X’, through Nessler’s reagent, a brown precipitate is obtained. The gas ‘X’ is
- $\text{H}_2\text{S}$
- $\text{CO}_2$
- $\text{NH}_3$
- $\text{Cl}_2$
The Correct Option is C
Approach Solution - 1
To answer the question of which gas 'X' produces a brown precipitate with Nessler's reagent, let's examine the options and the reaction involved.
Nessler's Reagent Reaction:
Nessler's reagent is a chemical solution used to detect the presence of ammonia (\(\text{NH}_3\)) in a sample. When ammonia gas is passed through Nessler's reagent, it reacts to form a brown precipitate of mercuric amido-iodide, which indicates the presence of ammonia.
The relevant reaction is:
\(\text{NH}_3 + 2\text{K}_2\text{HgI}_4 + 3\text{KOH} \rightarrow \text{HgO}\cdot\text{Hg(NH}_2\text{I}) + 7\text{KI} + 2\text{H}_2\text{O}\)
Option Analysis:
- \(\text{H}_2\text{S}\): When hydrogen sulfide is passed through Nessler's reagent, it does not form a brown precipitate. Instead, it reacts with metal ions to form black metal sulfides.
- \(\text{CO}_2\): Carbon dioxide does not react with Nessler's reagent to form any precipitate.
- \(\text{NH}_3\): This is the gas that reacts with Nessler's reagent to produce a brown precipitate, indicating the presence of ammonia.
- \(\text{Cl}_2\): Chlorine does not react with Nessler's reagent to produce a brown precipitate.
Therefore, the correct answer is the gas that reacts with Nessler's reagent to form a brown precipitate, which is \(\text{NH}_3\).
Conclusion:
Gas 'X' that forms a brown precipitate with Nessler's reagent is \(\text{NH}_3\), or ammonia.
Approach Solution -2
Nessler’s Reagent Reaction:
\(2K_2HgI_4 + NH_3 + 3KOH \rightarrow Hg_2(NH_3)_2I_2 + 7KI + 2H_2O\)
In the reaction above, the formation of a brown precipitate is characteristic of ammonia gas (\( NH_3 \)).
The Correct Answer is: \( NH_3 \)
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