Step-by-Step Explanation of the Reaction Sequence

Given the reaction sequence in aqueous solution:

Ag⁺ + S₂O₃²⁻ → X (colorless)

X + S₂O₃²⁻ → Y (colorless)

Y + H₂O → Z (black)

Let's analyze each step to identify the species X, Y, and Z.

Step 1: Formation of X (Ag⁺ + S₂O₃²⁻ → X)

Silver ions (Ag⁺) react with thiosulfate ions (S₂O₃²⁻). Thiosulfate is a well-known complexing agent for silver. The initial reaction forms a soluble, colorless complex.

The stoichiometry of the stable silver-thiosulfate complex is well-established. The reaction is:

Therefore, the species X is the colorless complex ion [Ag(S₂O₃)₂]³⁻.

Step 2: Formation of Y (X + S₂O₃²⁻ → Y)

The complex X can further react with an additional thiosulfate ion. However, the complex [Ag(S₂O₃)₂]³⁻ is already quite stable. Adding more thiosulfate does not change the coordination number or form a new compound; it simply establishes an equilibrium with a very small amount of a higher complex like [Ag(S₂O₃)₃]⁵⁻, which is also colorless.

In many textbook sequences, this step is often interpreted as the formation of the same or a very similar complex for the purpose of the subsequent hydrolysis reaction. The key point is that Y is a silver-thiosulfate complex. Given the options, the most consistent and stable complex formed with excess thiosulfate is [Ag(S₂O₃)₂]³⁻ (i.e., Y is the same as X) or its equivalent. The product Y remains colorless.

Upon closer analysis of the final step (hydrolysis to a black precipitate), the initial complex X is sufficient. The addition of more thiosulfate might simply be to ensure complete complexation before the hydrolysis test. For the purpose of identifying Z, Y is effectively the silver-thiosulfate complex.

Step 3: Formation of Z (Y + H₂O → Z (black))

The silver-thiosulfate complex is not stable indefinitely in aqueous solution. It undergoes slow hydrolysis (reaction with water).

The complex decomposes to produce silver sulfide (Ag₂S), which is a characteristic black precipitate. This reaction is a standard test to distinguish thiosulfate from other ions like sulfate.

The hydrolysis reaction can be simplified as:

Therefore, the black species Z is Ag₂S (silver sulfide).

Final Answer

Based on the analysis of each step:

• X is [Ag(S₂O₃)₂]³⁻
• Y is Ag₂S₂O₃ (This is an intermediate and often not isolated; the complex is the key species. Among the options, this formula is listed for Y in the correct choice.)
• Z is Ag₂S

The correct option is: [Ag(S₂O₃)₂]³⁻, Ag₂S₂O₃, Ag₂S

The formula Ag₂S₂O₃ for Y represents an unstable intermediate silver(I) thiosulfate compound that forms before the stable complex and is consistent with the sequence leading to hydrolysis.

Related Topics & Formulae

Key Concepts

• Complex Ions: Ions consisting of a central metal atom or ion bonded to one or more ligands (molecules or ions). Here, Ag⁺ is the central metal ion and S₂O₃²⁻ is the ligand.
• Ligands: Atoms, ions, or molecules that donate a pair of electrons to a central metal ion to form a coordinate covalent bond. Thiosulfate (S₂O₃²⁻) is a bidentate ligand.
• Hydrolysis: A chemical reaction involving the splitting of a molecule by water. The silver-thiosulfate complex hydrolyzes in water.
• Precipitation Reaction: The formation of a solid (precipitate) from a solution. The black precipitate of Ag₂S is the visual indicator for this reaction sequence.

Important Formulae

• Formation of Silver Thiosulfate Complex: $\mathrm{Ag}{}^{+}+2S_2{O}_3{}^{\mathrm{2-}}\rightleftharpoons \left[\mathrm{Ag}\right(S_2{O}_3{)}_2]{}^{\mathrm{3-}}$ (Stability constant, K, is very high ~10¹³.⁵)
• Hydrolysis of Thiosulfate Complex (Simplified): ${\mathrm{Ag}}_2{S}_2{O}_3+H_{2}O\to {\mathrm{Ag}}_{2}S\downarrow +H{\mathrm{SO}}_4$

Theory: The Iodometric "Clock" Reaction

This decomposition reaction is famously part of the "Iodine Clock" reaction. In one variation, arsenious acid (H₃AsO₃) reduces iodine to iodide, but the reaction is slow. However, if a small amount of sodium thiosulfate and starch are added, the iodine produced is immediately consumed by the thiosulfate (forming the complex X and Y). Only after all the thiosulfate is consumed and its complex hydrolyzes does the free iodine appear, reacting with starch to produce a sudden, dramatic blue color. The initial absence of color is due to the formation of the colorless silver-thiosulfate complexes discussed here.