Precipitation of Metal Ion with Ammoniacal Hydrogen Sulfide

In qualitative inorganic analysis, ammoniacal $\text{H}_2\text{S}$ (a solution of $\text{H}_2\text{S}$ in the presence of $\text{NH}_4\text{OH}$) provides an alkaline medium. The base $\text{NH}_4\text{OH}$ removes protons from the weak acid $\text{H}_2\text{S}$, significantly increasing the concentration of sulfide ions ($\text{S}^{2-}$):

$\text{H}_2\text{S} + 2\text{NH}_4\text{OH} \rightleftharpoons 2\text{NH}_4^+ + \text{S}^{2-} + 2\text{H}_2\text{O}$

Let us analyze the behavior of each metal ion upon treatment with ammoniacal $\text{H}_2\text{S}$:

1. $\text{Zn(II)}$: $\text{Zn}^{2+}$ belongs to Group IV of the qualitative analysis scheme. The elevated concentration of $\text{S}^{2-}$ in the ammoniacal medium exceeds the solubility product ($K_{\text{sp}}$) of zinc sulfide ($\text{ZnS}$). Consequently, it precipitates as a white sulfide: $\text{Zn}^{2+} + \text{S}^{2-} \longrightarrow \text{ZnS}\downarrow \text{ (white)}$

2. $\text{Al(III)}$: Aluminum sulfide ($\text{Al}_2\text{S}_3$) cannot exist in aqueous solution because it hydrolyzes completely to form aluminum hydroxide due to the strong polarization of the $\text{Al}^{3+}$ ion: $2\text{Al}^{3+} + 3\text{S}^{2-} + 6\text{H}_2\text{O} \longrightarrow 2\text{Al(OH)}_3\downarrow + 3\text{H}_2\text{S}$ Thus, $\text{Al(III)}$ precipitates as a hydroxide ($\text{Al(OH)}_3$), not as a sulfide.

3. $\text{Fe(III)}$: In an ammoniacal solution, $\text{Fe}^{3+}$ precipitates as ferric hydroxide $\text{Fe(OH)}_3$ due to its extremely low solubility product ($K_{\text{sp}} \approx 10^{-38}$). Moreover, $\text{Fe(III)}$ sulfide ($\text{Fe}_2\text{S}_3$) is unstable in aqueous solution; $\text{S}^{2-}$ acts as a reducing agent, reducing $\text{Fe}^{3+}$ to $\text{Fe}^{2+}$ with the deposition of sulfur: $2\text{Fe}^{3+} + \text{S}^{2-} \longrightarrow 2\text{Fe}^{2+} + \text{S}\downarrow$ Hence, $\text{Fe(III)}$ does not precipitate in the form of its own sulfide ($\text{Fe}_2\text{S}_3$).

4. $\text{Mg(II)}$: Magnesium sulfide ($\text{MgS}$) is soluble in water, and under ammoniacal conditions, magnesium remains in solution (belongs to Group VI). Therefore, no sulfide precipitate is formed.

Thus, only $\text{Zn(II)}$ precipitates out in the form of its sulfide ($\text{ZnS}$).

Correct Option: (D)