Upon treatment with ammoniacal $\mathrm{H}_2 \mathrm{~S}$, the metal ion that precipitates as a sulfide is
Select the correct option:
A
Fe (III)
B
AI (III)
C
$\mathrm{Mg}(\mathrm{II})$
D
Zn(II)
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Solution
Ammoniacal $\mathrm{H}_2 \mathrm{~S}$ is group reagent of fourth group cationic radicals. $\mathrm{Fe}^{3+} \& \mathrm{Al}^{3+}$ will precipitate $\mathrm{Fe}(\mathrm{OH})_3$ and $\mathrm{Al}(\mathrm{OH})_3$ respectively. Only Zinc will form white precipitate of ZnS.
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