On passing a gas, '$\mathrm{X}$', through Nessler's regent, a brown precipitate is obtained. The gas '$\mathrm{X}$' is
Solution
<p>Nessler's Reagent Reaction :</p>
<p>$$\underset{\text { (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text { (lodine of Millon's base } \\ \text { (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}$$</p>
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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