Calomel (${\text{Hg}}_2{\text{Cl}}_2$) on reaction with ammonium hydroxide gives:

Calomel reacts with ammonia to produce a black mixture. This is a disproportionation reaction:
$${\text{Hg}}_2{\text{Cl}}_2+2{\text{NH}}_3\to \text{Hg}+{\text{Hg(NH}}_2)\text{Cl}+{\text{NH}}_4\text{Cl}$$
The white precipitate is mercury(II) amido chloride (${\text{HgNH}}_2\text{Cl}$), but it appears black due to the finely divided metallic mercury ($\text{Hg}$) formed. The question asks for the compound formed, and ${\text{HgNH}}_2\text{Cl}$ is the standard chemical product identified in such tests.