When excess of `SnCl_(2)` is added to a soin of `HgCl_(2)` a white ppt turning grey is obtained the grey colour is due to the formation of

Options

A. `Hg_(2)Cl_(2)`
B. `SnCl_(4)`
C. `Sn`
D. `Hg`

Answer

To solve the problem, we need to analyze the reaction between stannous chloride (SnCl₂) and mercuric chloride (HgCl₂). Here’s a step-by-step breakdown of the reaction and the formation of the grey precipitate:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Stannous chloride (SnCl₂)
- Mercuric chloride (HgCl₂)

### Step 2: Write the Initial Reaction
When stannous chloride is added to mercuric chloride, the following reaction occurs:
\[ \text{SnCl}_2 + \text{HgCl}_2 \rightarrow \text{Hg}_2\text{Cl}_2 + \text{SnCl}_4 \]
In this reaction, mercuric chloride reacts with stannous chloride to form mercurous chloride (Hg₂Cl₂), which is a white precipitate, and stannic chloride (SnCl₄).

### Step 3: Identify the Excess Reagent
The problem states that excess stannous chloride is added. This means that after the formation of the white precipitate (Hg₂Cl₂), there is still stannous chloride present in the solution.

### Step 4: Reaction of Excess Stannous Chloride
The excess stannous chloride will react with the white precipitate (Hg₂Cl₂) as follows:
\[ \text{Hg}_2\text{Cl}_2 + 2\text{SnCl}_2 \rightarrow 2\text{Hg} + 2\text{SnCl}_4 \]
In this reaction, mercurous chloride (Hg₂Cl₂) is reduced by stannous chloride to form elemental mercury (Hg), which is grey in color.

### Step 5: Conclusion
The grey color observed in the solution is due to the formation of elemental mercury (Hg) from the reduction of the white precipitate (Hg₂Cl₂) by excess stannous chloride.

### Final Answer
The grey color is due to the formation of **Hg (elemental mercury)**.

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