Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

When dilute \(\mathrm{HCl}\) is added to a solution of substance (X) which is water soluble, a white precipitate \((\mathrm{Y})\) is formed. Treatment of \((\mathrm{Y})\) with \(\mathrm{NH}_{4} \mathrm{OH}\) turns it black (Z). Treatment of \((\mathrm{Y})\) with \(\mathrm{H}_{2} \mathrm{~S}\) also turns it black. Identify \((X)\) here? (a) \(\mathrm{HgO}\) (b) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{HgSO}_{4}\) (d) \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\)

Step-by-Step Solution

Answer

The substance \(X\) is \(\mathrm{Hg}_2\left(\mathrm{NO}_3\right)_2\).

Step 1: Substance X Reaction with HCl

Identify the process when substance \(X\) reacts with dilute \(\mathrm{HCl}\). The information says that a white precipitate \((\mathrm{Y})\) is formed. This indicates that \(X\) must be a compound that forms an insoluble chloride upon reaction with \(\mathrm{HCl}\). In the given options, \(\mathrm{Hg}_2^{2+}\) ions form \(\mathrm{Hg}_2\mathrm{Cl}_2\) (calomel), a white precipitate, when they react with \(\mathrm{HCl}\). From the choices, \(\mathrm{Hg}_2\left(\mathrm{NO}_3\right)_2\) is likely to react this way.

Step 2: Treatment of Y with Ammonium Hydroxide

Now examine the reaction of the white precipitate \((Y)\), presumably \(\mathrm{Hg}_2\mathrm{Cl}_2\), with \(\mathrm{NH}_4\mathrm{OH}\). \(\mathrm{Hg}_2\mathrm{Cl}_2\) reacts with \(\mathrm{NH}_4\mathrm{OH}\) to form black \(\mathrm{Hg}\) and \(\mathrm{NH}_4\mathrm{Cl}\), which fits the description given.

Step 3: Treatment of Y with Hydrogen Sulfide

Consider the reaction of \(Y\) (\(\mathrm{Hg}_2\mathrm{Cl}_2\)) with \(\mathrm{H}_2\mathrm{S}\). \(\mathrm{H}_2\mathrm{S}\) can convert \(\mathrm{Hg}_2\mathrm{Cl}_2\) into a black precipitate of \(\mathrm{HgS}\), which is consistent with the problem statement. This further supports that \(Y\) is \(\mathrm{Hg}_2\mathrm{Cl}_2\).

Step 4: Conclusion: Identify Substance X

Bringing together the reactions and information given, the correct compound \(X\) that matches the description in the problem is \(\mathrm{Hg}_2\left(\mathrm{NO}_3\right)_2\). This compound forms calomel \((\mathrm{Hg}_2\mathrm{Cl}_2)\) upon reaction with \(\mathrm{HCl}\) which turns black upon further reactions with \(\mathrm{NH}_4\mathrm{OH}\) and \(\mathrm{H}_2\mathrm{S}\).

Key Concepts

Inorganic Chemistry

Inorganic chemistry is the branch of chemistry concerned with inorganic compounds, which are compounds not primarily made of carbon. While organic chemistry deals with carbon-containing compounds, inorganic chemistry focuses on a vast array of compounds including metals, minerals, and organometallic compounds.
Studying inorganic chemistry involves understanding the properties and behavior of these non-organic compounds, which play crucial roles in many technological and biological processes.
Some fundamental areas in inorganic chemistry include:

• Crystal field theory: This theory explains the electronic structure of transition metal compounds using concepts like crystal field splitting and ligand field theory.
• Coordination compounds: These are compounds where central metal atoms/ions are bonded to surrounding molecules/ions, known as ligands. They can form colorful and complex geometries.
• Acid-base reactions: Inorganic chemistry analyzes differences between acid and base behavior in various non-organic contexts.

Applications range from medical therapies and industrial processes to environmental science, reflecting the extensive influence of inorganic chemistry in everyday life and industrial applications.

Precipitation Reactions

Precipitation reactions are chemical reactions where two soluble salts in aqueous solution react to form an insoluble solid, called the precipitate. These reactions are a type of double displacement reaction and are governed by the solubility rules of the reagents.
For example, when the compound \((X)\) reacts with \(\mathrm{HCl}\), it forms a white precipitate of \((Y)\). This white precipitate forms because one of the products is insoluble in water.
The basic steps involved in identifying whether a precipitate will form are:

• Predict the products: By examining the ions present in the solution, determine the possible products of the reaction.
• Check solubility rules: Use solubility rules to ascertain if any of the products are insoluble in water.
• Observe the precipitate: If one of the products is insoluble, it will form as a solid, which can often be observed as a change in the solution's appearance.

These reactions are useful in various fields, including analytical chemistry for detecting and quantifying specific ions in a solution.

Complex Ions

Complex ions are a fascinating and crucial area of study within inorganic chemistry. These ions consist of a central metal atom or ion surrounded by other molecules or ions known as ligands. This arrangement forms a complex, and when charged, it becomes a complex ion.
Complex ions are essential in various biochemical and industrial processes. In our scenario, the compound \(\mathrm{Hg}_2\left(\mathrm{NO}_3\right)_2\) forms a complex with chloride ions to create white calomel. Yet, under specific conditions, these complex ions can undergo further reactions like those observed with \(\mathrm{NH}_4\mathrm{OH}\) and \(\mathrm{H}_2\mathrm{S}\).
Here are a few points on the behavior of complex ions:

• Formation: Complex ions form through coordination bonds, which occur when ligands donate electron pairs to the metal center.
• Geometry: The arrangement of the ligands around the metal ion can lead to different geometric structures such as linear, square planar, or tetrahedral.
• Stability: The stability of complex ions is influenced by the types of ligands and the overall charge of the ion.

Understanding complex ions can help elucidate many chemical processes, from biological enzyme functions to manufacturing catalysts.