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In Nessler's reagent, the ion present is (a) \(\mathrm{HgI}_{4}^{2-}\) (b) \(\mathrm{Hg} \mathrm{I}_{2}^{2-}\) (c) \(\mathrm{Hg}^{+}\) (d) \(\mathrm{Hg}^{2+}\)

Short Answer

Expert verified
In Nessler's reagent, the ion present is (a) 6HgI_4^{2-}7.

Step by step solution

01

Introduction to Nessler's Reagent

Nessler's reagent is used in chemical tests to detect the presence of ammonia. It is a solution that typically contains mercury and iodine compounds.
02

Understanding the Composition of Nessler's Reagent

Nessler's reagent is formed by the reaction of mercury(II) iodide (6HgI_27) and potassium iodide in alkaline medium. The mixture forms a complex ion that is responsible for its utility in ammonia detection.
03

Identifying the Ion in Nessler's Reagent

Among the options given, the ion 6HgI_4^{2-}7 is formed in Nessler's reagent. This tetrathiomercurate complex coordinates with ammonia causing a color change which is used in various analyses.
04

Selecting the Correct Answer

From the choices given, we determine that (a) 6HgI_4^{2-}7 is the correct ion present in Nessler's reagent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Tests for Ammonia
Ammonia detection is a critical aspect of various chemical and environmental applications. One effective way to test for ammonia is by using Nessler's reagent, a classical reagent for ammonia detection in solution. Here's how it works: - **Color Change**: When Nessler's reagent interacts with ammonia, a color change occurs. This change can be from colorless to pale yellow or even orange-brown, indicating the presence of ammonia. - **Ease of Detection**: This colorimetric method is straightforward, allowing even small concentrations of ammonia to be detected without sophisticated equipment. The color shift results from the formation of a complex with the components of Nessler's reagent. This simplicity and reliability make it a favored choice in educational and professional laboratories for ammonia quantification.
Mercury and Iodine Compounds
Nessler's reagent is made from interesting chemistry involving mercury and iodine compounds. Here’s a breakdown of what happens:- **Mercury(II) Iodide**: This is one of the key active compounds in Nessler's reagent, initially formed by mixing mercury and iodine.- **Complex Formation**: When mixed with potassium iodide in an alkaline medium, this compound associates with iodine to form the ion \[ \text{HgI}_4^{2-} \].- **Safety Considerations**: Mercury compounds are toxic and should be handled with care. Proper lab equipment like gloves and goggles must be worn to avoid direct exposure.Understanding these components is crucial because they form the active complex that reacts with ammonia, aiding in its detection.
Complex Ion Formation
The magic of Nessler's reagent lies in its ability to form a stable complex ion. Here is how the complex ion formation takes place:- **Alkaline Environment**: In an alkaline environment, mercury interacts with iodide ions, forming the \[ \text{HgI}_4^{2-} \] ion.- **Reaction with Ammonia**: When ammonia enters the solution, it interacts with this complex ion. Ammonia acts as a ligand, binding with the mercury ion to further stabilize the complex.- **Visible Outcome**: The binding of ammonia to this ion leads to a visible color change. This is primarily due to the structural shift in the complex, amplifying the ability to detect ammonia.Such ion formation not only helps detect ammonia but also introduces students to the broader concepts of complex chemistry, showcasing how metal ions can interact with various ligands to produce diverse outcomes.

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Most popular questions from this chapter

For which of the following metals may be hydrometallurgical extractive method be employed? (a) copper (b) iron (c) chromium (d) tin

In the process of extraction of gold Roasted gold ore \(+\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{O}_{2}}{\longrightarrow}[\mathrm{X}]+\mathrm{OH}^{-}\) \([\mathrm{X}]+\mathrm{Zn} \longrightarrow[\mathrm{Y}]+\mathrm{Au}\) Identify the complexes of \([\mathrm{X}]\) and \([\mathrm{Y}]\) (a) \(\mathrm{Y}=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}\) (b) \(\mathrm{X}=\left[\mathrm{Au}(\mathrm{CN})_{4}\right]^{3-}\) (c) \(\mathrm{X}=\left[\mathrm{Au}(\mathrm{CN})_{2}\right]\) (d) \(\mathrm{Y}=\left[\mathrm{Zn}(\mathrm{CN})_{6}\right]^{4-}\)

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