Question

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

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

Answer 1

Answer: (A)

$HgO.N{H}_2\left(Hg\right)Cl$

When calomel ($H{g}_{2}C{l}_2$) reacts with $N{H}_{4}OH$ solution, the compound formed is $[HgO.Hg(N{H}_2\left)Cl\right]$. It is black in colour.
$H{g}_{2}C{l}_2+2N{H}_{4}OH\to [HgO.Hg(N{H}_2\left)Cl\right]\downarrow +N{H}_{4}Cl+2H_{2}O$