Question

H₂​S is passed through an acidified solution of Ag, Cu and Zn. Which forms precipitate?
(A) $Ag$
(B) $Zn$
(C) $Cu$
(D) None of these

Answer 1

Hint: Cations are grouped on the basis of selective precipitation of sparingly soluble salts. They are categorised into $6$ groups along with zero group cations that fall in no category. Group $2$ are precipitated in acid medium in the form of sulphides.

Complete Step by Step Solution:
group $2$ is divided into further two categories.
Group $2$A cations include $H{g^{ + 2}},P{b^{ + 2}},B{i^{ + 3}},C{u^{ + 2}},C{d^{ + 2}}$. They are called the copper group.

Group $2B$ cations include $S{b^{ + 3}},S{b^{ + 5}},A{s^{ + 3}},A{s^{ + 5}},S{n^{ + 2}},S{n^{ + 4}}$. They are called arsenic groups.

These two groups are based on their solubility in alkaline yellow ammonium sulphide.
Group $2$A is insoluble in alkaline yellow ammonium sulphide whereas group $2$B is soluble in alkaline yellow ammonium sulphide. Copper groups have low electronegativity values and thus, do not dissolve in alkaline medium. Whereas the arsenic group has high electronegativity values hence they dissolve in the solution of alkaline ammonium sulphide to give thio anions.

So, in acidic medium when ${H_2}S$ is passed, either copper group precipitates or arsenic group. Silver and zinc do not belong to the group $2$. Hence, they will not get precipitated. $C{u^{ + 2}}$ reacts with ${H_2}S$ to form black precipitate $CuS$.
Thus, the correct option is C.

Note: Cadmium is sometimes missed in groups . In the presence of excess chloride ions (from hydrochloric acid in the first group), cadmium forms a soluble complex ${[CdC{l_4}]^{2 - }}$. Due to this the concentration of cadmium decreases and it is not sufficient enough to precipitate cadmium sulphide on passing ${H_2}S$in acidic condition.