Match List-I with List-II.
List-I (Colloid preparation method)
A. Hydrolysis
B. Reduction
C. Oxidati

Q: Why is the formation of sulphur sol considered oxidation?

In the reaction between $\text{SO}_2$ and $\text{H}_2\text{S}$, the hydrogen sulphide ($\text{H}_2\text{S}$) is oxidized to elemental sulphur ($\text{S}$) by losing hydrogen.

Q: What exactly is double decomposition?

It is a type of reaction where two reactant compounds exchange their ions to form two entirely new compounds. In our case, $\text{As}_2\text{O}_3$ and $\text{H}_2\text{S}$ exchange ions to form $\text{As}_2\text{S}_3$ and $\text{H}_2\text{O}$.

Q: Why is formaldehyde used in the preparation of gold sol?

Formaldehyde ($\text{HCHO}$) acts as a mild reducing agent. It reduces the $\text{Au}^{3+}$ ions in $\text{AuCl}_3$ to neutral $\text{Au}$ atoms, which then aggregate to form colloidal particles.

Q: Are these colloidal sols lyophilic or lyophobic?

The sols prepared by these chemical methods (like metal sols, sulphide sols, and hydroxide sols) are generally lyophobic (solvent-hating) in nature and require stabilizing agents.

Q: Can we use other reducing agents for preparing gold sol?

Yes! Besides formaldehyde, other reducing agents like stannous chloride ($\text{SnCl}_2$), hydrazine, or even tannic acid can be used to reduce gold chloride to gold sol.

Question

Match List-I with List-II.

$\begin{array}{ll} 	extbf{List-I (Colloid preparation method)} & 	extbf{List-II (Chemical reaction)} \ 	ext{A. Hydrolysis} & 	ext{1. } 2	ext{AuCl}_3 + 3	ext{HCHO} + 3	ext{H}_2	ext{O} ightarrow 2	ext{Au(sol)} + 3	ext{HCOOH} + 6	ext{HCl} \ 	ext{B. Reduction} & 	ext{2. } 	ext{As}_2	ext{O}_3 + 3	ext{H}_2	ext{S} ightarrow 	ext{As}_2	ext{S}_3	ext{(sol)} + 3	ext{H}_2	ext{O} \ 	ext{C. Oxidation} & 	ext{3. } 	ext{SO}_2 + 2	ext{H}_2	ext{S} ightarrow 3	ext{S(sol)} + 2	ext{H}_2	ext{O} \ 	ext{D. Double Decomposition} & 	ext{4. } 	ext{FeCl}_3 + 3	ext{H}_2	ext{O} ightarrow 	ext{Fe(OH)}_3	ext{(sol)} + 3	ext{HCl} \end{array}$

Choose the most appropriate answer from the options given below.

Options:

(A) A $ightarrow$ 1, B $ightarrow$ 3, C $ightarrow$ 2, D $ightarrow$ 4

(B) A $ightarrow$ 4, B $ightarrow$ 1, C $ightarrow$ 3, D $ightarrow$ 2

(C) A $ightarrow$ 4, B $ightarrow$ 2, C $ightarrow$ 3, D $ightarrow$ 1

(D) A $ightarrow$ 1, B $ightarrow$ 2, C $ightarrow$ 4, D $ightarrow$ 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we dive into the world of surface chemistry, one of the most fascinating topics is the preparation of colloidal solutions. Colloids aren't just simple mixtures; they require specific chemical methods to ensure the particles fall right into that magical $1$ to $1000 	ext{ nm}$ range.

In this problem, we are given four distinct chemical methods: **Hydrolysis**, **Reduction**, **Oxidation**, and **Double Decomposition**. Our mission is to match these methods with their corresponding chemical reactions. Let's break them down one by one.

### The Master Reactions

**1. Hydrolysis**
Hydrolysis involves the breaking of chemical bonds by the addition of water. When we look at the given reactions, the reaction of Ferric chloride ($	ext{FeCl}_3$) with water perfectly fits this description:

$$ 	ext{FeCl}_3 + 3	ext{H}_2	ext{O} ightarrow 	ext{Fe(OH)}_3	ext{(sol)} + 3	ext{HCl} $$

Here, Ferric chloride undergoes hydrolysis to form a positively charged sol of Ferric hydroxide. Thus, **A matches with 4**.

**2. Reduction**
Reduction is the gain of electrons or a decrease in oxidation state. In the preparation of metal sols like gold, silver, or platinum, we use reducing agents. Looking at our list, the reaction involving Gold chloride ($	ext{AuCl}_3$) and formaldehyde ($	ext{HCHO}$) is a classic example:

$$ 2	ext{AuCl}_3 + 3	ext{HCHO} + 3	ext{H}_2	ext{O} ightarrow 2	ext{Au(sol)} + 3	ext{HCOOH} + 6	ext{HCl} $$

Gold is reduced from a $+3$ oxidation state to $0$, forming a beautiful purple-red gold sol. Therefore, **B matches with 1**.

**3. Oxidation**
Oxidation involves the loss of electrons or an increase in oxidation state. Non-metal sols like sulphur are often prepared by oxidation. When Hydrogen sulphide ($	ext{H}_2	ext{S}$) gas is bubbled through a solution of Sulphur dioxide ($	ext{SO}_2$), the $	ext{H}_2	ext{S}$ is oxidized to elemental sulphur:

$$ 	ext{SO}_2 + 2	ext{H}_2	ext{S} ightarrow 3	ext{S(sol)} + 2	ext{H}_2	ext{O} $$

This gives us a colloidal sol of sulphur. Hence, **C matches with 3**.

**4. Double Decomposition**
Double decomposition is a reaction where two compounds exchange ions to form two new compounds. The reaction between Arsenious oxide ($	ext{As}_2	ext{O}_3$) and Hydrogen sulphide ($	ext{H}_2	ext{S}$) is a textbook example:

$$ 	ext{As}_2	ext{O}_3 + 3	ext{H}_2	ext{S} ightarrow 	ext{As}_2	ext{S}_3	ext{(sol)} + 3	ext{H}_2	ext{O} $$

The exchange of ions leads to the formation of a negatively charged Arsenious sulphide sol. So, **D matches with 2**.

### Final Conclusion

By carefully analyzing the chemical nature of each reaction, we have successfully decoded the matching sequence. The correct mapping is **A $ightarrow$ 4, B $ightarrow$ 1, C $ightarrow$ 3, and D $ightarrow$ 2**. This is a high-yield concept for JEE, so make sure these standard preparation reactions are at your fingertips!

Select Answer:

Visualized Solution

Methods of Colloid Preparation

Hydrolysis of $FeCl_{3}$

Reduction of $AuCl_{3}$

Oxidation of $H_{2} S$

Double Decomposition

Final Matching

The Sigma Insight: Colloids, Micelles and and Emulsions

Analyzing the Setup

The Master Reactions

Final Conclusion

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