Question

Which of the following is soluble in yellow ammonium sulphide?
(A) $CuS$
(B) $CdS$
(C) $SnS$
(D) $PbS$

Answer 1

Hint: In qualitative analysis of cations, solubility in alkaline yellow ammonium sulphide is used to divide group $2$ into two groups namely $2$A and $2$B. Group $2$A is called the copper group and Group $2$B is called the arsenic group.

Complete Step by Step Solution:
Group $2$ cations are precipitated on passing ${H_2}S$ in presence of acidic medium. The acidic medium is used to suppress ${H_2}S$ which is a weak acid so that sulphide ion concentration does not exceed and the group $4$is not precipitated out.

Group $2$ is divided into two categories: Group $2$A and Group $2$B.
Group $2$A cations are called as copper group and they include$H{g^{ + 2}},P{b^{ + 2}},B{i^{ + 3}},C{u^{ + 2}},C{d^{ + 2}}$. Group $2$A is insoluble in alkaline yellow ammonium sulphide. Reason is that the Copper group has low electronegativity values and thus, do not dissolve in alkaline medium.

Group $2B$cations are called as arsenic group and they include $S{b^{ + 3}},S{b^{ + 5}},A{s^{ + 3}},A{s^{ + 5}},S{n^{ + 2}},S{n^{ + 4}}$. Group $2$B is soluble in alkaline yellow ammonium sulphide because the arsenic group has high electronegativity values hence, they dissolve in the solution of alkaline ammonium sulphide to give thio anions.

Hence, $SnS$ belongs to group $2$ B which is soluble in alkaline yellow ammonium sulphide whereas $CuS$ ,$CdS$ and $PbS$ belongs to group $2$A which is insoluble in alkaline yellow ammonium sulphide.
Hence, the correct answer is B.

Note: Sodium sulphide or ${(N{H_4})_2}S$ cannot be used, rather ${(N{H_4})_2}{S_x}$ is used because it serves as the basis of separation. If ${(N{H_4})_2}S$ is used instead of ${(N{H_4})_2}{S_x}$, tin ($II$) sulphide is insoluble and thus, will not dissolve in colourless ammonium sulphide, ${(N{H_4})_2}S$. To dissolve $SnS$, ammonium polysulphide is required which partly acts as an oxidising agent.