Problem 118 - A Complete Resource Book in Chemistry for JEE Main

Question

What are the products obtained respectively when KI solution in excess is added to \(\mathrm{HgCl}_{2}\) and \(\mathrm{CuSO}_{4}\) solution separately? (a) \(\mathrm{HgI}_{2}, \mathrm{~K}_{3} \mathrm{CuI}_{4}\) (b) \(\mathrm{K}_{2} \mathrm{HgI}_{4}, \mathrm{CuI}\) (c) \(\mathrm{K}_{2} \mathrm{HgI}_{4}, \mathrm{~K}_{2} \mathrm{CuI}_{4}\) (d) \(\mathrm{Hg}_{2} \mathrm{I}_{2}, \mathrm{Cu}_{2} \mathrm{I}_{2}\)

Step-by-Step Solution

Answer

The products are K2HgI4 for HgCl2 and CuI for CuSO4, matching (b).

Step 1: Reaction of HgCl2 with KI

When an excess KI solution is added to HgCl2, an initial reaction occurs:\[ \text{HgCl}_2 + 2\text{KI} \rightarrow \text{HgI}_2 + 2\text{KCl} \]Initially, mercuric iodide (HgI2) forms as a yellow precipitate. However, in excess KI, this precipitate continues to dissolve forming a soluble complex.\[ \text{HgI}_2 + 2\text{KI} \rightarrow \text{K}_2\text{HgI}_4 \]Thus, the final product when excess KI is used is \( \text{K}_2\text{HgI}_4 \).

Step 2: Reaction of CuSO4 with KI

Now, consider the reaction between CuSO4 and excess KI. First, a displacement reaction occurs:\[ \text{CuSO}_4 + 2\text{KI} \rightarrow \text{CuI}_2 + \text{K}_2\text{SO}_4 \]However, since CuI2 is unstable, it immediately decomposes into CuI and I2:\[ 2\text{CuI}_2 \rightarrow 2\text{CuI} + \text{I}_2 \]Therefore, excess KI ensures that \(\text{CuI}\) is the final insoluble product.

Key Concepts

Precipitation Reactions

In inorganic chemistry, a precipitation reaction occurs when two aqueous solutions react to form an insoluble solid, known as a precipitate. These reactions are straightforward and primarily involve ionic compounds exchanging partners. In the context of the exercise, when mercury(II) chloride (\(\text{HgCl}_2\)) reacts with potassium iodide (\(\text{KI}\)), a yellow precipitate of mercuric iodide (\(\text{HgI}_2\)) forms initially. This precipitate occurs as the iodide ions replace chloride ions.

• Most precipitation reactions involve ionic compounds.
• They are commonly used to synthesize new compounds or to purify products.
• The solubility of the product determines whether a precipitate will form.

Precipitate formation plays a crucial role in chemical analysis and separation techniques. The yellow appearance of mercuric iodide indicates an immediate reaction and the formation of a distinct solid from the solution.

Complex Formation

Complex formation occurs when simple ions or molecules bind together to form a more complex entity, often increasing solubility. After the initial formation of mercuric iodide (\(\text{HgI}_2\)) in the presence of excess potassium iodide (\(\text{KI}\)), the precipitate transforms. The excess iodide ions bond further with mercuric iodide to form potassium tetraiodomercurate(II) (\(\text{K}_2\text{HgI}_4\)), which is soluble.

• Complex formation helps dissolve initially insoluble products.
• Such transformations exploit the coordinative properties of ions.
• It often involves transition metals forming stable structures with non-metals.

This process is essential in many biochemical systems and industrial applications where increased solubility is necessary. In this reaction's context, the transition from \(\text{HgI}_2\) to \(\text{K}_2\text{HgI}_4\) exemplifies how complexation can change the chemical properties of a precipitate.

Displacement Reactions

Displacement reactions, or single replacement reactions, occur when an element replaces another element in a compound. These reactions are pivotal in the transformation of ionic compounds. In the exercise, copper(II) sulfate (\(\text{CuSO}_4\)) and potassium iodide (\(\text{KI}\)) participate in such a reaction. Potassium iodide donates iodine to replace sulfate in copper sulfate, forming copper(I) iodide (\(\text{CuI}_2\)) and potassium sulfate (\(\text{K}_2\text{SO}_4\)).

• These reactions often feature metals displacing other cationic species.
• Displacement reactions are integral in metallic corrosion and extraction.
• Products can often change state or become new solids or gases.

Displacement reactions not only allow for the creation of new compounds but also can be used to separate or purify materials in various chemical processes. In our scenario, the unstable \(\text{CuI}_2\) eventually decomposes, forming copper(I) iodide (\(\text{CuI}\)) and iodine.

Reactions with Halides

Halides, compounds consisting of a halogen bonded to a more electropositive element, often undergo interesting chemical reactions. In the reactions from the exercise, iodide acting as the halide reacts distinctly, depending on the counterpart. With \(\text{HgCl}_2\), iodide precipitates and then dissolves further to form a complex. Meanwhile, with \(\text{CuSO}_4\), iodide participates first in a displacement reaction, subsequently leading to a different set of products.

• Halides typically react via substitution or, like in these cases, precipitation and displacement.
• The reactive nature of halides is due to their high electronegativity.
• Outcomes depend heavily on the specific halide and other reactive species involved.

Reactions involving halides are widespread in both organic and inorganic chemistry. They play roles from altering molecular structure to revitalizing materials through exchanges, significantly impacting the course of chemical processes and reactions.