Upon treatment with ammoniacal ${{H}_{2}}S$, the metal ion that is precipitated as a sulfide is:
a.) Fe (III)
b.) Al (III)
c.) Mg (II)
d.) Zn (II)

Hint: A precipitation reaction is a type of reaction in which two soluble salts in a fluid solution mix with each other and one of them is insoluble salt called precipitate. The precipitate stays in the solution as a barrier and it can also be removed from the liquid by using the process of filtration or decantation etc.

Complete step by step answer:
Upon the treatment with ammoniacal ${{H}_{2}}S$, the metal ion which precipitates as a sulfide is Zn (II). The chemical reaction involved is shown below:
\[Zn\text{ }\left( II \right)+{{H}_{2}}S(ammoniacal)\to ZnS+2H\]

The ammonium hydroxide and ${{H}_{2}}S$ gas is the group reagent for the Zn (II) only this can form the precipitate with ammoniacal ${{H}_{2}}S$. ZnS forms a white precipitate.
Whereas Al (III) form hydroxide on reaction with ammonical hydrogen sulphide, similarly Fe (III) also form hydroxide on reaction with ammonical hydrogen sulphide.
Mg (II) form respective bicarbonate when it is treated with ammoniacal hydrogen sulphide.
So, the correct answer is “Option D”.

Additional Information:
When ZnS is heated in the presence of excess of air around 1200K ZnO is formed. This process is known as roasting. The reaction involves is shown below:
\[2ZnS+3{{O}_{2}}\xrightarrow{\Delta (heating)}2ZnO+2S{{O}_{2}}\]

Zinc oxide formed acts as an amphoteric oxide and it reacts with acids to form zinc salt. Zinc oxide reacts with a strong base to form zincates, when this is strongly heated it assumes a yellow color which disappears when it is cooled down.

Note: Zinc sulfite mainly occurs in the minerals known as sphalerite and wurtzite, both of these are intrinsic wide-band gap semiconductors. Zinc sulfite is commonly used as a white pigment it helps in the detection of alpha rays, which render its luminescent.