Question

A precipitate of which of the following would be obtained when \[HCl\] is added to a solution of stannous sulphide (\[SnS\] ) in yellow ammonium sulphide?
A. \[SnS\]
B. \[Sn{S_2}\]
C. \[{\left( {N{H_4}} \right)_2}Sn{S_2}\]
D. \[S{n_2}{S_3}\]

Answer 1

Hint: The cation of tin (\[S{n^{2 + }}\] ,\[S{n^{4 + }}\]) belongs to Group II in the qualitative inorganic analysis scheme. Group II cations form sulphides that are not soluble in acids.

Complete Step by Step Solution:
In qualitative inorganic analysis, we analyse a given salt or a given mixture of salts to determine their elemental composition. This is usually done through the detection of their constituent ions in aqueous solutions. Firstly, the given salt sample must be brought into an aqueous solution. Then, various reagents are used to treat the solution and test for certain reactions characteristic to the ions present. These characteristic reactions involve either a colour change, precipitate formation, or other visible changes. The analysis is conducted separately for anions and cations.

For the qualitative inorganic analysis, anions and cations are usually classified into groups based on their properties. The groups mentioned here are different from the groups in the Periodic Table.

The cations that are analysed in the qualitative inorganic analysis are divided into six groups. Each group has a common reagent which can be used to separate the cation from its solution.

The salt in question is stannous sulphide (\[SnS\]) which has a divalent cation of tin\[S{n^{2 + }}\].\[S{n^{2 + }}\]belongs in the 2nd analytical group of cations along with\[S{n^{4 + }}\]cation. The 2nd analytical group consists of cations that form sulphides which are insoluble in acids. The common reagent can be any substance that gives sulphide ions (\[{S^{2 - }}\] ) in aqueous solutions. This test with sulphide ions must be conducted in the presence of dilute hydrochloric acid (\[HCl\]).

Ammonia’s action is useful in distinguishing the cations. Sulphides of\[S{n^{2 + }}\]and\[S{n^{4 + }}\]cations are soluble in yellow ammonium sulphide (\[{\left( {N{H_4}} \right)_2}{S_X}\] ) because they form polysulphide complexes.
\[SnS(aq) + \left( {N{H_4}} \right){}_2{S_2}(aq) \to \left( {N{H_4}} \right){}_2Sn{S_3}(aq) \overset{HCl}{\rightarrow} Sn{S_2}(s) + \left( {N{H_4}} \right){}_2S\] where \[Sn{S_2}\] is called Tin (IV) sulphide which precipitates out as a golden yellow precipitate.
Thus, the correct option is B.

Note: The 2^nd group test with sulphides is conducted in the presence of dilute\[HCl\] to keep the sulphide concentration low enough to allow the precipitation of 2nd group cations alone. Without diluting \[HCl\], cations of the 4th group may precipitate early leading to misleading results.