`Hg_(2)l_(2)darr ("green") underset("with" H_(2)O) overset("boiled") to` products
Which of the following statement is correct with respect to the products ?

Options

A. Black precipitate of mercury `(I)` oxide is formed.
B. Voilet colour gas is evolved.
C. Red precipitate of `Hgl_(2)` is formed.
D. Mercury is obtained.

Answer

To solve the problem regarding the reaction of Hg₂I₂ (mercury(I) iodide) when boiled with water, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant is mercury(I) iodide, represented as Hg₂I₂, which is green in color.

2. **Reaction with Water**:
When Hg₂I₂ is boiled with water (H₂O), it undergoes a chemical reaction.

3. **Products Formation**:
The reaction leads to the formation of two products:
- Mercury(II) iodide (HgI₂), which is red in color.
- Elemental mercury (Hg), which may appear as a precipitate.

4. **Change in Ion States**:
In the reactant Hg₂I₂, the cation is Hg₂²⁺. Upon reaction, it changes to Hg²⁺ in HgI₂ and elemental Hg.

5. **Color Change**:
Initially, the solution is green due to Hg₂I₂, and after the reaction, it turns red due to the formation of HgI₂.

6. **Conclusion**:
Therefore, the correct statement regarding the products formed when Hg₂I₂ is boiled with water is that it produces red HgI₂ and elemental mercury (Hg) as a precipitate.