Question

When calomel react with $N{H}_{4}OH$ solution the compound formed is:

• A. $N{H}_2-Hg-Cl$
• B. $H{g}_{2}C{l}_{2}N{H}_3$
• C. $Hg(N{H}_3{)}_{2}C{l}_2$
• D. $HgC{l}_{2}N{H}_3$

Answer 1

The correct option is A $N{H}_2-Hg-Cl$

Calomel electrode consists of Pt wire along with paste of $Hg$, $H{g}_{2}C{l}_2$ and saturated $KCl$.

Its reaction with $N{H}_{4}OH$ can be given as follow:

$H{g}_{2}C{l}_2+2N{H}_{4}OH\to \underset{\text{Black mixture}}{Hg+Hg\left(N{H}_2\right)}Cl+N{H}_{4}Cl+2H_{2}O$