Chemistry, asked by kunju3495, 1 year ago

Upon treatment with ammoniacal H₂S, the metal ion that precipitates as a sulfide is (a) Fe(III) (b) Al(III) (c) Mg(II) (d)Zn(II)

Answers

Answered by priya9993
4
Option d is correct.. Becz, group reagant of Zn(ll) is ammonical H2S...
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Answered by GulabLachman
0

Answer:

d) Zn(II)

Explanation:

The basic reaction of ammoniacal H2S is also called as group reagent.

As we know that in the periodic table the Sulphides present in the 4th group are insoluble in water because of its solubility product(Ksp) . So these gives precipitates when H2S reacts with NH4OH.

Now we see that Zn(II) lies in the same group so the option d) is correct.

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