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

F e (I I I)

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A l (I I I)

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M g (I I)

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Z n (I I)

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Solution

The correct option is B $Z n (I I)$
Upon treatment with ammonical $H_{2} S$ , the metal ion that precipitates as a sulfide is $Z n (I I)$ . The chemical equation for this is as follows:
$Z n (I I) + H_{2} S (a m m o n i c a l) ⟶ Z n S + 2 H$
As, $N H_{4} O H + H_{2} S$ gas, is the group reagent for the $Z n (I I)$ ion. So, only this can give the precipitate with ammonical $H_{2} S$ .

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Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is:

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