On passing a gas, 'X', through Nessler's reagent, a brown precipitate is obtained. The gas 'X' is

Step 1: Recall Nessler's Reagent

Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II) ($\ce{K2HgI4}$ in $\ce{KOH}$ or $\ce{NaOH}$).

Step 2: Consider the reaction with specific gases

Nessler's reagent is the standard diagnostic test for the presence of ammonia ($\ce{NH3}$) gas or ammonium ($\ce{NH4+}$) ions. When ammonia gas is passed through Nessler's reagent, a brown or yellow-brown precipitate forms.

$\ce{2K2HgI4 + NH3 + 3KOH -> HgO.Hg(NH2)I v + 7KI + 2H2O}$

This brown precipitate is known as the iodide of Millon's base.

Step 3: Eliminate other options

• $\ce{H2S}$ gas typically forms black precipitates with lead or dark colors with metals, but not specifically targeted by Nessler's reagent in this manner.
• $\ce{CO2}$ turns lime water milky, but has no particular colored reaction with Nessler's reagent.
• $\ce{Cl2}$ is a strong oxidizing agent, but the brown precipitate definition with Nessler's is exclusively for ammonia.

Step 4: Conclusion

The gas X must be ammonia ($\ce{NH3}$).

Key Points to Remember:

• Nessler's Reagent = Alkaline $\ce{K2HgI4}$.
• Its primary use is detecting $\ce{NH4+}$ ions or $\ce{NH3}$ gas.
• The identifying result is a brown precipitate called iodide of Millon's base.