Problem 35 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The pair(s) of ions where BOTH the ions are precipitated upon passing \(\mathrm{H}_{2} \mathrm{~S}\) gas in presence of dilute \(\mathrm{HCl}\), is (are) (a) \(\mathrm{Ba}^{2+}, \mathrm{Zn}^{2+}\) (b) \(\mathrm{Bi}^{3+}, \mathrm{Fe}^{3+}\) (c) \(\mathrm{Cu}^{2+}, \mathrm{Pb}^{2+}\) (d) \(\mathrm{Hg}^{2+}, \mathrm{Bi}^{3+}\)

Step-by-Step Solution

Answer

The pairs are (c) \( \text{Cu}^{2+}, \text{Pb}^{2+} \) and (d) \( \text{Hg}^{2+}, \text{Bi}^{3+} \).

Step 1: Understand the Conditions

Passing \( \text{H}_2\text{S} \) gas in the presence of dilute \( \text{HCl} \) is used to precipitate out ions as sulfides. Under these conditions, only less soluble sulfides of metal ions will precipitate, specifically those of the Group 2 of qualitative analysis including ions like \( \text{Cu}^{2+}, \text{Pb}^{2+}, \text{Hg}^{2+}, \text{Bi}^{3+} \).

Step 2: Recall the Solubility Rules

In the presence of dilute \( \text{HCl} \), Group 2 metal sulfides (e.g., \( \text{Cu}^{2+}, \text{Pb}^{2+}, \text{Hg}^{2+}, \text{Bi}^{3+} \)) are precipitated. These ions form very insoluble sulfides and therefore precipitate out upon passing \( \text{H}_2\text{S} \) gas.

Step 3: Check Each Pair of Ions

Consider which pairs of ions can precipitate under these specific conditions: - (a) \( \text{Ba}^{2+} \) does not precipitate as a sulfide in dilute \( \text{HCl} \).- (b) \( \text{Fe}^{3+} \) does not precipitate as a sulfide in the presence of dilute \( \text{HCl} \).- (c) Both \( \text{Cu}^{2+} \) and \( \text{Pb}^{2+} \) precipitate as sulfides.- (d) Both \( \text{Hg}^{2+} \) and \( \text{Bi}^{3+} \) precipitate as sulfides.

Step 4: Conclusion

From our evaluation, we determine that pairs (c) \( \text{Cu}^{2+}, \text{Pb}^{2+} \) and (d) \( \text{Hg}^{2+}, \text{Bi}^{3+} \) will result in the precipitation of both ions when \( \text{H}_2\text{S} \) gas is passed in dilute \( \text{HCl} \).

Key Concepts

Group 2 Metal Sulfides

When studying qualitative analysis, Group 2 metal sulfides refer to certain metal ions that form insoluble sulfide compounds. In the context of the exercise, these are the ions like \( \text{Cu}^{2+} \), \( \text{Pb}^{2+} \), \( \text{Hg}^{2+} \), and \( \text{Bi}^{3+} \). These ions are known to readily precipitate as sulfides under specific conditions. The role of these sulfides is crucial in qualitative inorganic analysis, as they assist in identifying specific ions in a mixture by forming visible precipitates. These metal ions belong to a broader classification in qualitative analysis and are not the same as the Group 2 elements on the periodic table (alkaline earth metals like calcium and magnesium). Instead, they form a specific group based on their similar chemical behavior in forming sulfides.

• These sulfides are typically very insoluble in water.
• The precipitation helps indicate the presence of these metal ions in a solution.

Precipitation Reactions

A precipitation reaction is a process where two substances in solution form a solid known as a precipitate. For the reaction described in the exercise, \( \text{H}_2\text{S} \) gas is passed through a solution containing different metal ions in the presence of dilute \( \text{HCl} \). This addition can cause the formation of a solid sulfide if the metal ion present tends to form an insoluble sulfide. These reactions are important for separating and identifying ions, known as qualitative analysis.The two key factors influencing precipitation include:

• The solubility of the resulting compound.
• The ionic strength of the solution.

Understanding precipitation reactions is crucial in predicting which compounds will form the precipitate in a reaction process. This exercise focuses on reactions where the ions form insoluble sulfide precipitates, such as \( \text{Cu}^{2+}, \text{Pb}^{2+}, \text{Hg}^{2+}, \text{Bi}^{3+} \) when specific conditions are met.

Solubility Rules

Solubility rules are guidelines used to predict whether a compound will dissolve in water, forming a homogeneous solution, or remain insoluble, leading to precipitation. In the exercise, these rules help us know under which circumstances a metal sulfide will precipitate upon adding \( \text{H}_2\text{S} \) gas in dilute \( \text{HCl} \).The pivotal rule here is: many metal sulfides are insoluble, especially those of lower solubility in water. When \( \text{H}_2\text{S} \) gas is bubbled through an acidic solution, such as one containing dilute \( \text{HCl} \), insoluble sulfides precipitate out.

• Metal ions like \( \text{Cu}^{2+}, \text{Pb}^{2+}, \text{Hg}^{2+}, \text{Bi}^{3+} \) form poorly soluble sulfides, making them precipitate.
• Metals whose sulfides are highly soluble do not precipitate and thus remain in solution.

Being aware of solubility rules allows chemists to correctly predict and make use of what will occur during precipitation, helping them to isolate and analyze various ions in a mixture effectively.