`SnCl_(2)` gives a precipitate with a solution of `HgCl_(2)` In this process `HgCl_(2)` is :

Explanation

To determine the role of `HgCl₂` in the reaction with `SnCl₂`, we can analyze the reaction step by step.

### Step 1: Write the reaction
The reaction between `SnCl₂` and `HgCl₂` can be represented as:
\[ \text{SnCl}_2 + \text{HgCl}_2 \rightarrow \text{SnCl}_4 + \text{Hg}_2\text{Cl}_2 \]

### Step 2: Identify the oxidation states
- In `SnCl₂`, Sn has an oxidation state of +2.
- In `HgCl₂`, Hg has an oxidation state of +2.
- In `SnCl₄`, Sn has an oxidation state of +4.
- In `Hg₂Cl₂`, the oxidation state of Hg is +1 (since there are two Hg atoms sharing the +2 charge).

### Step 3: Determine the changes in oxidation states
- Sn goes from +2 in `SnCl₂` to +4 in `SnCl₄`, which means Sn is oxidized.
- Hg goes from +2 in `HgCl₂` to +1 in `Hg₂Cl₂`, which means Hg is reduced.

### Step 4: Identify the oxidizing and reducing agents
- The substance that gets reduced (Hg in this case) is the oxidizing agent.
- The substance that gets oxidized (Sn in this case) is the reducing agent.

### Conclusion
In this reaction, `HgCl₂` acts as an oxidizing agent because it undergoes reduction from +2 to +1 oxidation state.

### Final Answer
`HgCl₂` is an oxidizing agent.

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