Question

Upon treatment with ammonical $H_{2}S$, the metal ion that precipitates as a sulphide is:

• A. ${Fe}^{3+}$
• B. ${Al}^{3+}$
• C. ${Mg}^{2+}$
• D. ${Zn}^{2+}$

Answer 1

The correct option is D ${Zn}^{2+}$

$ZnC{l}_2+H_{2}S\to 2HCl+ZnS$

In $ZnC{l}_2$, $Zn$ looses two electrons and form $Z{n}^{2+}$ and reacts with sulfur of $H_{2}S$ and forms $ZnS$ which is a precipitate.

Hence option D is correct.