Key idea: group separation with ammoniacal $H_2S$

In qualitative inorganic analysis, passing $H_2S$ through an ammoniacal / basic medium increases the concentration of $S^{2-}$ ions because: $H_2S \rightleftharpoons H^+ + HS^- \rightleftharpoons 2H^+ + S^{2-}$ In basic medium, $H^+$ is removed, so the equilibrium shifts right and more $S^{2-}$ is produced.

Therefore, those metal ions whose sulphides have very low solubility product precipitate as sulphides in ammoniacal medium.

Examine each option

A. $Fe^{3+}$
$Fe^{3+}$ does not normally precipitate as sulphide with ammoniacal $H_2S$. In basic/ammoniacal medium, iron is generally precipitated as hydroxide: $Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3 \downarrow$ So this is not the required ion.

B. $Al^{3+}$
Aluminium does not precipitate as sulphide because $Al_2S_3$ is hydrolysed strongly in water. In ammoniacal medium, aluminium is precipitated as hydroxide: $Al^{3+} + 3OH^- \rightarrow Al(OH)_3 \downarrow$ So not correct.

C. $Mg^{2+}$
$MgS$ is sufficiently soluble / not precipitated under these conditions. Magnesium is not precipitated by ammoniacal $H_2S$ as sulphide. So not correct.

D. $Zn^{2+}$
Zinc belongs to the group of cations that precipitate as sulphides in ammoniacal medium: $Zn^{2+} + S^{2-} \rightarrow ZnS \downarrow$ Hence $Zn^{2+}$ precipitates as zinc sulphide.

Conclusion

The metal ion that precipitates as a sulphide upon treatment with ammoniacal $H_2S$ is: $\boxed{Zn^{2+}}$

Comparison with stored answer

Stored correct answer = D
Derived answer = D

Therefore, the derived answer agrees with the stored answer.