Question

\(\mathrm{Q}\) and \(\mathrm{S}\) can be distinguished by (A) \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\) (B) Na-metal (C) \(\mathrm{Cu}_{2} \mathrm{Cl}_{2} / \mathrm{NH}_{4} \mathrm{OH}\) (D) \(\mathrm{AgNO}_{3} / \mathrm{NH}_{4} \mathrm{OH}\)

Short answer

The best reagent to distinguish between halogens \(\mathrm{Q}\) and \(\mathrm{S}\) is \(\boxed{\text{(D) } \mathrm{AgNO}_{3} / \mathrm{NH}_{4} \mathrm{OH}}\). Silver nitrate reacts with halide ions to form silver halide precipitates, which have different colors and solubility properties, allowing for easy distinction between different halogens.

Step-by-step solution

Recall the properties of halogens

To solve this exercise, we first need to recall the properties of halogens. Halogens are elements found in Group 17 of the periodic table and include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Each halogen has distinct chemical and physical properties, which allow them to be differentiated using specific reagents.

Option A: \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\)

Bromine water (\(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\)) can be used as a test for unsaturated hydrocarbons, like alkenes or alkynes. However, it cannot be used to distinguish between different halogens.

Option B: Na-metal

Sodium metal can react with halogens, forming different sodium halides. However, the resulting compounds have similar physical properties and reactivity towards other reagents, making it difficult to distinguish between different halogens.

Option C: \(\mathrm{Cu}_{2} \mathrm{Cl}_{2} / \mathrm{NH}_{4} \mathrm{OH}\)

This is the reagent used for the Finkelstein reaction. Sodium iodide (\(\mathrm{NaI}\)) can be used to test for the presence of iodine in a given sample. When \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\) is added to ammonium hydroxide (\(\mathrm{NH}_{4} \mathrm{OH}\)), Finkelstein's reagent is formed. This reagent, upon reaction with \(\mathrm{NaI}\), forms a precipitate of CuI which indicates the presence of iodine. However, since the Finkelstein reaction is specific for iodine, it won't be able to distinguish between other halogens.

Option D: \(\mathrm{AgNO}_{3} / \mathrm{NH}_{4} \mathrm{OH}\)

This reagent consists of silver nitrate (\(\mathrm{AgNO}_{3}\)) in ammonium hydroxide (\(\mathrm{NH}_{4} \mathrm{OH}\)). Silver nitrate reacts with halide ions (\(\mathrm{Cl}^{-}\), \(\mathrm{Br}^{-}\), \(\mathrm{I}^{-}\)) to form silver halide precipitates, which have different colors and solubility properties: - Silver chloride (AgCl) - white precipitate, soluble in ammonium hydroxide - Silver bromide (AgBr) - pale yellow precipitate, slightly soluble in ammonium hydroxide - Silver iodide (AgI) - yellow precipitate, insoluble in ammonium hydroxide By observing the color of the precipitate and its solubility in ammonium hydroxide, we can easily distinguish between different halogens. In conclusion, the best reagent to distinguish between halogens \(\mathrm{Q}\) and \(\mathrm{S}\) is: \(\boxed{\text{(D) } \mathrm{AgNO}_{3} / \mathrm{NH}_{4} \mathrm{OH}}\)

Key concepts

Finkelstein Reaction

The Finkelstein reaction provides a clear method for identifying the presence of iodide in organic compounds. It is a substitution reaction where an alkyl halide is treated with sodium iodide in acetone, leading to the exchange of the halide ion for iodide.

The reaction takes advantage of the fact that sodium iodide is soluble in acetone, whereas sodium chloride and sodium bromide are not. If a compound containing chlorine or bromine is used in the reaction, the newly formed sodium chloride or sodium bromide precipitates out of the reaction mixture. In contrast, if iodine is present, sodium iodide remains in solution.

In this way, the Finkelstein reaction can confirm the presence of iodine when there is no precipitate. However, this specificity also means it cannot differentiate between chlorine and bromine, as both will result in a precipitate.

Silver Nitrate Halide Test

The silver nitrate halide test is an exceptionally useful tool in qualitative analysis to identify the presence of halide ions in a sample. When a solution containing halide ions is treated with silver nitrate, a silver halide precipitate is formed. This precipitate varies in color based on the halide present:

• Silver chloride (AgCl) forms a white precipitate that is soluble in dilute ammonia, giving a clear solution upon the addition of ammonium hydroxide.
• Silver bromide (AgBr) appears as a pale yellow precipitate that is slightly soluble in dilute ammonia, resulting in a partially clear solution.
• Silver iodide (AgI) produces a distinct yellow precipitate that is insoluble in dilute ammonia, leaving the precipitate unchanged.

By observing these distinct reactions, the silver nitrate halide test not only confirms the presence of halogens but also allows for their differentiation, making it an invaluable analytical technique in organic chemistry.

Properties of Halogens

Halogens are a fascinating group of elements located in Group 17 of the periodic table, which includes fluorine, chlorine, bromine, iodine, and astatine. Each halogen has distinct properties that are reflected in their physical states at room temperature:

• Fluorine (F) and chlorine (Cl) are gases, with chlorine having a heavier molecular weight.
• Bromine (Br) is the only liquid halogen at room temperature.
• Iodine (I) and astatine (At) are both solids, with iodine displaying a characteristic shiny, metallic lustre.

These physical properties accompany a wide range of chemical behaviors, such as their electron affinities and reactivities. Fluorine, being the most electronegative element, is extremely reactive, whereas astatine, the rarest natural halogen, is the least reactive. Understanding these properties allows chemists to predict the reactivity and selectivity of halogens within different chemical contexts, making the identification and differentiation of halogens a fundamental skill in organic chemistry.