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

Options

A. reduced
B. oxidised
C. converted into a complex compound containing both Sn and Hg
D. converted into a chloro complex of Hg.

Answer

To solve the question regarding the reaction between SnCl₂ and HgCl₂, we need to analyze the changes in oxidation states of the elements involved in the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the balanced chemical equation
The reaction between tin(II) chloride (SnCl₂) and mercury(II) chloride (HgCl₂) can be represented as follows:

\[ \text{SnCl}_2 + \text{HgCl}_2 \rightarrow \text{SnCl}_4 + \text{Hg}_2\text{Cl}_2 \]

### Step 2: Determine the oxidation states
Next, we need to determine the oxidation states of tin (Sn) and mercury (Hg) in the reactants and products.

- In SnCl₂:
- Chlorine (Cl) has an oxidation state of -1.
- Let the oxidation state of Sn be \( x \).
- The equation becomes: \( x + 2(-1) = 0 \) → \( x = +2 \).
- Therefore, Sn in SnCl₂ is in the +2 oxidation state.

- In HgCl₂:
- Let the oxidation state of Hg be \( y \).
- The equation becomes: \( y + 2(-1) = 0 \) → \( y = +2 \).
- Therefore, Hg in HgCl₂ is also in the +2 oxidation state.

- In SnCl₄:
- Let the oxidation state of Sn be \( a \).
- The equation becomes: \( a + 4(-1) = 0 \) → \( a = +4 \).
- Therefore, Sn in SnCl₄ is in the +4 oxidation state.

- In Hg₂Cl₂:
- Let the oxidation state of Hg be \( b \).
- The equation becomes: \( 2b + 2(-1) = 0 \) → \( 2b - 2 = 0 \) → \( b = +1 \).
- Therefore, Hg in Hg₂Cl₂ is in the +1 oxidation state.

### Step 3: Identify oxidation and reduction
Now, we can analyze the changes in oxidation states:

- **Tin (Sn)**: Changes from +2 in SnCl₂ to +4 in SnCl₄. This is an increase in oxidation state, indicating that Sn is oxidized.
- **Mercury (Hg)**: Changes from +2 in HgCl₂ to +1 in Hg₂Cl₂. This is a decrease in oxidation state, indicating that Hg is reduced.

### Step 4: Conclusion
Since HgCl₂ undergoes a reduction process (gaining electrons), the correct answer to the question is that HgCl₂ is **reduced**.

### Final Answer
In this process, HgCl₂ is **reduced**.

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