Problem 53 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An aqueous solution containing one mole of \(\mathrm{HgI}_{2}\) and two moles of NaI is orange in colour. On addition of excess \(\mathrm{NaI}\), the solution becomes colouress. The orange colour reappears on subsequent addition of \(\mathrm{NaOCl}\). Explain with equations.

Step-by-Step Solution

Answer

The color change is due to the formation and dissociation of \( [HgI_4]^{2-} \) with iodide ions being oxidized by \( \mathrm{NaOCl} \).

Step 1: Understanding Initial Reactions

When one mole of \( \mathrm{HgI}_{2} \) is combined with two moles of \( \mathrm{NaI} \) in aqueous solution, an orange complex is formed due to the interaction of \( \mathrm{HgI}_{2} \) with \( \mathrm{I^{-}} \) ions to form \( \mathrm{HgI_4^{2-}} \). The equation for this reaction is: \[ \mathrm{HgI_2 + 2I^{-} \rightarrow [HgI_4]^{2-}} (orange) \].

Step 2: Addition of Excess NaI

Adding excess \( \mathrm{NaI} \) shifts the equilibrium towards the formation of \( \mathrm{[HgI_4]^{2-}} \), which remains in solution. The excess iodide ions cause the solution to become colorless as more iodide ions stabilize the \( \mathrm{[HgI_4]^{2-}} \) complex.

Step 3: Explanation of Colorless Solution

In the presence of excess iodide ions, the orange color is not visible because the complex formation stabilizes the colorless \( \mathrm{[HgI_4]^{2-}} \) species, suppressing free \( \mathrm{HgI_2} \) which shows the orange hue.

Step 4: Adding NaOCl

When \( \mathrm{NaOCl} \) is added to the complex solution, it acts as an oxidizing agent. It oxidizes \( \mathrm{I^{-}} \) ions to \( \mathrm{I_2} \), thus displacing the equilibrium back to liberate \( \mathrm{HgI_2} \): \[ 2[HgI_4]^{2-} + \mathrm{NaOCl} \rightarrow \mathrm{HgI_2 (solid)} + \mathrm{Cl^-} +2 \mathrm{I_2} + \mathrm{Na^+} \]. The resurgence of \( \mathrm{HgI_2} \), being orange in color, causes the solution to regain its orange hue.

Key Concepts

HgI2 complex

In coordination chemistry, a complex compound forms when a central metal atom or ion binds with surrounding molecules or ions known as ligands. One such interesting example is the formation of a complex with mercuric iodide (HgI2). Typically, HgI2 is an orange solid; however, when it interacts with iodide ions (I\(^-\)), it forms a complex ion which changes its behavior. When HgI2 reacts with two iodide ions from NaI in an aqueous solution, a complex ion called tetraiodomercurate(II), or \([HgI_4]^{2-}\), is formed. This stabilizes the mercury ion by surrounding it with four iodide ions. The equation for this process is: \[\mathrm{HgI_2 + 2I^- \rightarrow [HgI_4]^{2-}} \]This compound is particularly interesting because it is orange, yet in solution with excess iodide ions, it becomes colorless. This is due to the complex being reduced in energy, which affects its color properties, making the solution appear colorless under certain conditions.

NaI and NaOCl reactions

The reactions involving sodium iodide (NaI) and sodium hypochlorite (NaOCl) showcase the dynamic behavior of iodide ions under different chemical conditions. Adding excess NaI to the initially orange solution containing \(\mathrm{[HgI_4]^{2-}}\) continues to push the equilibrium towards more complex formation. Because the \(\mathrm{[HgI_4]^{2-}}\) is colorless due to its stabilized form, adding even more iodide ions makes the solution colorless as the free orangery \(\mathrm{HgI_2}\) is no longer present.On the other hand, introducing NaOCl into the already colorless solution containing \(\mathrm{[HgI_4]^{2-}}\) triggers a fascinating change. Sodium hypochlorite acts as an oxidizing agent, which uniquely impacts this chemical system. It causes the iodide ions (\(\mathrm{I^-}\)) present in the solution to oxidize into iodine (\(\mathrm{I_2}\)), thus shifting the chemical equilibrium. This displacement reaction then converts the colorless \(\mathrm{[HgI_4]^{2-}}\) back to orange \(\mathrm{HgI_2}\), rendering the solution orange again: \[2[HgI_4]^{2-} + \mathrm{NaOCl} \rightarrow \mathrm{HgI_2 (solid)} + \mathrm{Cl^-} + 2 \mathrm{I_2} + \mathrm{Na^+} \]

oxidizing agents in chemical reactions

Oxidizing agents are substances that can accept electrons from other molecules in a chemical reaction, causing those molecules to be oxidized. These agents are crucial in various reactions for facilitating electron transfer, thus changing the oxidation state of substances involved.In the discussed reaction, NaOCl is an oxidizing agent. Oxidizing agents generally have a strong affinity for electrons, allowing them to draw electrons from other substances. In the solution containing the \(\mathrm{[HgI_4]^{2-}}\) complex, NaOCl oxidizes iodide ions (\(\mathrm{I^-}\)) to iodine (\(\mathrm{I_2}\)). This reaction reduces chlorine from NaOCl to chloride ions (\(\mathrm{Cl^-}\)), while NaOCl's active component handles electron withdrawal from iodide:

• The iodide ions lose electrons to become iodine.
• The chlorine part of NaOCl gains those electrons, reducing it to chloride ions.

This dynamic change illustrates how oxidizing agents like NaOCl can effectively shift chemical equilibria, therefore playing a pivotal role in reactions involving transformation of spectator ions and complexes.