Question

When ${H_2}S$ is passed through II group sometimes the solution becomes milky. It indicates the presence of:
A. Acidic salt
B. An oxidising agent
C. Thiosulfate
D. A reducing agent

Answer 1

Hint: The 2nd analytical group of cations consists of ions which form acid-insoluble sulfides. Cations in the II group include:\[C{d^{2 + }},\;B{i^{3 + }},\;C{u^{2 + }},\;A{s^{3 + }},{\text{ }}A{s^{5 + }},\;S{b^{3 + }},{\text{ }}S{b^{5 + }},{\text{ }}S{n^{2 + }},{\text{ }}S{n^{4 + }}\;and{\text{ }}H{g^{2 + }}\] .\[P{b^{2 + }}\] is usually also included here in addition to the first group. Although these methods refer to solutions that contain sulfide (\[{S^{2 - }}\] ), these solutions actually only contain \[{H_2}S\] and bisulfide (\[H{S^ - }\] ). Sulfide (\[{S^{2 - }}\] ) does not exist in appreciable concentrations in water.

Complete step by step answer:
The reagent used can be any substance that gives \[{S^{2 - }}\] ions in such solutions; most commonly used are hydrogen sulfide (at \[0.2 - 0.3{\text{ }}M\] ). The addition of hydrogen sulfide can often prove to be a lumber-some process and therefore sodium sulfide can also serve the purpose. The test with the sulfide ion must be conducted in the presence of dilute HCl. Its purpose is to keep the sulfide ion concentration at a required minimum, so as to allow the precipitation of 2nd group cations alone.
When ${H_2}S$ is passed through II group, sometimes the solution becomes milky. It indicates the presence of any oxidising agent in the solution that oxidizes sulfur from $ - 2$ in \[{H_2}S\] to $ + 4$ in $S{O_2}$ . The formation of sulfur dioxide and sulfate ions in the solution will turn the solution milky.

Thus, the correct option is B. An oxidizing agent .

Note:
If dilute acid is not used in the reaction of hydrogen sulfide, the early precipitation of 4th group cations (if present in solution) may occur, thus leading to misleading results. Acids beside \[HCl\] are rarely used. Sulfuric acid may lead to the precipitation of the 5th group cations, whereas nitric acid oxidizes the sulfide ion in the reagent, forming colloidal sulfur.