Problem 2 - A Complete Resource Book in Chemistry for JEE Main

Question

On passing \(\mathrm{H}_{2} \mathrm{~S}\) gas in group II sometimes the solution turns milky. It indicates the presence of (a) acidic salt (b) thiosulphate (c) oxidizing agent (d) reducing agent

Step-by-Step Solution

Answer

(b) thiosulphate

Step 1: Understand the Reaction

When \(\mathrm{H}_2\mathrm{S}\) gas is passed through the group II cation solution, it generally forms a precipitate of metal sulfides such as \(\mathrm{PbS}\) or \(\mathrm{HgS}\), which are insoluble. However, the presence of certain anions can affect this process.

Step 2: Analyze the Milky Appearance

A milky appearance when passing \(\mathrm{H}_2\mathrm{S}\) suggests that something in the solution is reacting to inhibit the usual formation of a clear precipitate.

Step 3: Consider Possible Inhibitors

Certain ions can react with the sulfide ions, preventing their combination with metal ions to form regular precipitates. Thiosulfate (\(\mathrm{S}_2\mathrm{O}_3^{2-}\)) is known to react with metal ions, forming soluble complexes, leading to a milky suspension rather than a solid precipitate.

Step 4: Conclude with Correct Option

The presence of thiosulfate (\(\mathrm{S}_2\mathrm{O}_3^{2-}\)) explains the milky suspension since it reacts to form soluble complexes with the sulfide ions rather than typical insoluble metal sulfides.

Key Concepts

H2S gas reactions

In chemistry, hydrogen sulfide or \( \mathrm{H}_2\mathrm{S} \) gas is a notable player when it comes to reactions involving metal ions. This gas is characterized by its rotten egg smell and is a weak acid that can donate hydrogen ions. When \( \mathrm{H}_2\mathrm{S} \) is bubbled through solutions containing metal ions, particularly from group II of the periodic table, it readily forms metal sulfides.
These metal sulfides are typically insoluble and precipitate out of the solution, often resulting in a characteristic color or appearance, such as black for lead sulfide \( \mathrm{PbS} \) or yellow for cadmium sulfide \( \mathrm{CdS} \). This reaction occurs because the sulfide ions from \( \mathrm{H}_2\mathrm{S} \) combine with the metal ions to create these solid metal sulfide compounds.

• The reaction is: \( \mathrm{Metal}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{Metal~Sulfide~(precipitate)} \).
• Precipitation is usually a sign that the \( \mathrm{H}_2\mathrm{S} \) gas is forming an insoluble compound with the cations in the solution.

Understanding \( \mathrm{H}_2\mathrm{S} \) gas reactions is essential because they are used in analytical chemistry to detect and quantify the presence of specific metal ions.

thiosulfate complex formation

Thiosulfate ions (\( \mathrm{S}_2\mathrm{O}_3^{2-} \)) play an intriguing role in altering typical metal ion reactions. When they are present in a solution where \( \mathrm{H}_2\mathrm{S} \) is being passed, they can create unexpected changes. The presence of these ions leads to the formation of soluble complexes with metal ions, which disrupts the usual precipitation process.
This interaction can be visualized as an alternative pathway where metal ions bind with thiosulfate instead of sulfide ions from \( \mathrm{H}_2\mathrm{S} \). Rather than precipitating, these complexes stay dissolved, resulting in a milky appearance of the solution due to the dispersion of these complexes throughout the liquid.

• Thiosulfate ions (\( \mathrm{S}_2\mathrm{O}_3^{2-} \)) form complexes via an interaction known as complexation.
• This can be represented by the equation: \( \mathrm{Metal}^{2+} + \mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow [\mathrm{Metal}(\mathrm{S}_2\mathrm{O}_3)] \text{complex}\).

Understanding thiosulfate complex formation is crucial because it explains why typical reactions are altered and why suspensions might appear rather than precipitates when \( \mathrm{H}_2\mathrm{S} \) gas is introduced.

metal sulfide precipitation

One of the key outcomes when \( \mathrm{H}_2\mathrm{S} \) interacts with metal ions is the precipitation of metal sulfides. Precipitation is when a solid forms in a solution as a result of a chemical reaction. For many metal ions, especially those in group II like \( \mathrm{Pb}^{2+} \) or \( \mathrm{Cd}^{2+} \), this reaction results in an insoluble sulfide.Once formed, these metal sulfides are usually clearly visible as precipitates, often floating in the liquid or settling at the bottom.

• The chemical process might begin with \( \mathrm{H}_2\mathrm{S} \) dissociating in solution to release \( \mathrm{S}^{2-} \) ions.
• \( \mathrm{S}^{2-} \) ions then combine with metal ions \( \mathrm{M}^{2+} \) in the solution: \( \mathrm{M}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{MS}\).
• The resulting compound is a metal sulfide, a solid that is often insoluble, cloudy, or colored, which distinguishes it from the solution.

Metal sulfide precipitation is an important concept in the analysis of metal ions, providing a visual cue for chemical reactions and an understanding of solubility dynamics in inorganic chemistry.