Preparation of gold sol by below method is bassed on
`2AuCI_(3)+3HCHO+3H_(2)Orarr 2Au(sol ) + 3HCOOH + 6HCI `

Options

A. Hydrolysis
B. Double decomposition
C. Reduction
D. Oxidation

Answer

To prepare gold sol using the given reaction:

\[ 2 \text{AuCl}_3 + 3 \text{HCHO} + 3 \text{H}_2\text{O} \rightarrow 2 \text{Au (sol)} + 3 \text{HCOOH} + 6 \text{HCl} \]

we can break down the process step by step:

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Gold(III) chloride (\( \text{AuCl}_3 \))
- Formaldehyde (\( \text{HCHO} \))
- Water (\( \text{H}_2\text{O} \))

### Step 2: Understand the Reaction
In this reaction, gold(III) chloride is being reduced to form gold sol. The formaldehyde acts as a reducing agent.

### Step 3: Reduction Process
- The oxidation state of gold in \( \text{AuCl}_3 \) is +3.
- When it is reduced to gold sol (\( \text{Au} \)), the oxidation state of gold becomes 0.
- This indicates that gold is gaining electrons, which is characteristic of a reduction reaction.

### Step 4: Products of the Reaction
The products formed from the reaction are:
- Gold sol (\( \text{Au (sol)} \))
- Formic acid (\( \text{HCOOH} \))
- Hydrochloric acid (\( \text{HCl} \))

### Step 5: Conclusion on the Method
The preparation of gold sol by this method is based on a reduction process. The formaldehyde reduces the gold(III) ions to metallic gold.

### Final Answer
The preparation of gold sol by the given method is based on **reduction**.

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